{"id":1877,"date":"2018-01-11T20:55:49","date_gmt":"2018-01-11T20:55:49","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/derivatives-of-exponential-and-logarithmic-functions\/"},"modified":"2018-05-23T18:52:35","modified_gmt":"2018-05-23T18:52:35","slug":"derivatives-of-exponential-and-logarithmic-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/derivatives-of-exponential-and-logarithmic-functions\/","title":{"raw":"3.9 Derivatives of Exponential and Logarithmic Functions","rendered":"3.9 Derivatives of Exponential and Logarithmic Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Find the derivative of exponential functions.<\/li>\r\n \t<li>Find the derivative of logarithmic functions.<\/li>\r\n \t<li>Use logarithmic differentiation to determine the derivative of a function.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1169738221270\">So far, we have learned how to differentiate a variety of functions, including trigonometric, inverse, and implicit functions. In this section, we explore derivatives of exponential and logarithmic functions. As we discussed in <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/introduction\/\">Introduction to Functions and Graphs<\/a>, exponential functions play an important role in modeling population growth and the decay of radioactive materials. Logarithmic functions can help rescale large quantities and are particularly helpful for rewriting complicated expressions.<\/p>\r\n\r\n<div id=\"fs-id1169737795671\" class=\"bc-section section\">\r\n<h1>Derivative of the Exponential Function<\/h1>\r\n<p id=\"fs-id1169738221359\">Just as when we found the derivatives of other functions, we can find the derivatives of exponential and logarithmic functions using formulas. As we develop these formulas, we need to make certain basic assumptions. The proofs that these assumptions hold are beyond the scope of this course.<\/p>\r\n<p id=\"fs-id1169738045927\">First of all, we begin with the assumption that the function [latex]B(x)=b^x, \\, b&gt;0[\/latex], is defined for every real number and is continuous. In previous courses, the values of exponential functions for all rational numbers were defined\u2014beginning with the definition of [latex]b^n[\/latex], where [latex]n[\/latex] is a positive integer\u2014as the product of [latex]b[\/latex] multiplied by itself [latex]n[\/latex] times. Later, we defined [latex]b^0=1, \\, b^{\u2212n}=\\frac{1}{b^n}[\/latex] for a positive integer [latex]n[\/latex], and [latex]b^{s\/t}=(\\sqrt[t]{b})^s[\/latex] for positive integers [latex]s[\/latex] and [latex]t[\/latex]. These definitions leave open the question of the value of [latex]b^r[\/latex] where [latex]r[\/latex] is an arbitrary real number. By assuming the <em>continuity<\/em> of [latex]B(x)=b^x, \\, b&gt;0[\/latex], we may interpret [latex]b^r[\/latex] as [latex]\\underset{x\\to r}{\\lim}b^x[\/latex] where the values of [latex]x[\/latex] as we take the limit are rational. For example, we may view [latex]{4}^{\\pi}[\/latex] as the number satisfying<\/p>\r\n\r\n<div id=\"fs-id1169738212703\" class=\"equation unnumbered\">[latex]\\begin{array}{l}4^3&lt;4^{\\pi}&lt;4^4, \\, 4^{3.1}&lt;4^{\\pi}&lt;4^{3.2}, \\, 4^{3.14}&lt;4^{\\pi}&lt;4^{3.15},\\\\ 4^{3.141}&lt;4^{\\pi}&lt;4^{3.142}, \\, 4^{3.1415}&lt;4^{\\pi}&lt;4^{3.1416}, \\, \\cdots \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1169737765653\">As we see in the following table, [latex]4^{\\pi}\\approx 77.88[\/latex].<\/p>\r\n\r\n<table id=\"fs-id1169738222068\" summary=\"This table has seven rows and four columns. The first row is a header row and it labels each column. The first column header is x, the second column header is 4x, the third column header is x, and the fourth column header is 4x. Under the first column are the values 43, 43.1, 43.14, 43.141, 43.1415. Under the second column are the values 64, 73.5166947198, 77.7084726013, 77.8162741237, 77.8702309526, 77.8799471543. Under the third column are the values 43.141593, 43.1416, 43.142, 43.15, 43.2, and 44. Under the fourth column are the values 77.8802710486, 77.8810268071, 77.9242251944, 78.7932424541, 84.4485062895, and 256.\"><caption>Approximating a Value of [latex]4^{\\pi}[\/latex]<\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>[latex]x[\/latex]<\/th>\r\n<th>[latex]4^x[\/latex]<\/th>\r\n<th>[latex]x[\/latex]<\/th>\r\n<th>[latex]4^x[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex]4^3[\/latex]<\/td>\r\n<td>64<\/td>\r\n<td>[latex]4^{3.141593}[\/latex]<\/td>\r\n<td>77.8802710486<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]4^{3.1}[\/latex]<\/td>\r\n<td>73.5166947198<\/td>\r\n<td>[latex]4^{3.1416}[\/latex]<\/td>\r\n<td>77.8810268071<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]4^{3.14}[\/latex]<\/td>\r\n<td>77.7084726013<\/td>\r\n<td>[latex]4^{3.142}[\/latex]<\/td>\r\n<td>77.9242251944<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]4^{3.141}[\/latex]<\/td>\r\n<td>77.8162741237<\/td>\r\n<td>[latex]4^{3.15}[\/latex]<\/td>\r\n<td>78.7932424541<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]4^{3.1415}[\/latex]<\/td>\r\n<td>77.8702309526<\/td>\r\n<td>[latex]4^{3.2}[\/latex]<\/td>\r\n<td>84.4485062895<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]4^{3.14159}[\/latex]<\/td>\r\n<td>77.8799471543<\/td>\r\n<td>[latex]4^4[\/latex]<\/td>\r\n<td>256<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1169737707453\">We also assume that for [latex]B(x)=b^x, \\, b&gt;0[\/latex], the value [latex]B^{\\prime}(0)[\/latex] of the derivative exists. In this section, we show that by making this one additional assumption, it is possible to prove that the function [latex]B(x)[\/latex] is differentiable everywhere.<\/p>\r\n<p id=\"fs-id1169737728646\">We make one final assumption: that there is a unique value of [latex]b&gt;0[\/latex] for which [latex]B^{\\prime}(0)=1[\/latex]. We define [latex]e[\/latex] to be this unique value, as we did in <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/introduction\/\">Introduction to Functions and Graphs<\/a>. <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_03_09_001\">(Figure)<\/a> provides graphs of the functions [latex]y=2^x, \\, y=3^x, \\, y=2.7^x[\/latex], and [latex]y=2.8^x[\/latex]. A visual estimate of the slopes of the tangent lines to these functions at 0 provides evidence that the value of [latex]e[\/latex] lies somewhere between 2.7 and 2.8. The function [latex]E(x)=e^x[\/latex] is called the <strong>natural exponential function<\/strong>. Its inverse, [latex]L(x)=\\log_e x=\\ln x[\/latex] is called the <strong>natural logarithmic function<\/strong>.<\/p>\r\n\r\n<div id=\"CNX_Calc_Figure_03_09_001\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205529\/CNX_Calc_Figure_03_09_001.jpg\" alt=\"The graphs of 3x, 2.8x, 2.7x, and 2x are shown. In quadrant I, their order from least to greatest is 2x, 2.7x, 2.8x, and 3x. In quadrant II, this order is reversed. All cross the y-axis at (0, 1).\" width=\"731\" height=\"592\" \/> <strong>Figure 1.<\/strong> The graph of [latex]E(x)=e^x[\/latex] is between [latex]y=2^x[\/latex] and [latex]y=3^x[\/latex].[\/caption]<\/div>\r\n<div class=\"wp-caption-text\"><\/div>\r\n<p id=\"fs-id1169737717318\">For a better estimate of [latex]e[\/latex], we may construct a table of estimates of [latex]B^{\\prime}(0)[\/latex] for functions of the form [latex]B(x)=b^x[\/latex]. Before doing this, recall that<\/p>\r\n\r\n<div id=\"fs-id1169738116003\" class=\"equation unnumbered\">[latex]B^{\\prime}(0)=\\underset{x\\to 0}{\\lim}\\frac{b^x-b^0}{x-0}=\\underset{x\\to 0}{\\lim}\\frac{b^x-1}{x} \\approx \\frac{b^x-1}{x}[\/latex]<\/div>\r\n<p id=\"fs-id1169737790576\">for values of [latex]x[\/latex] very close to zero. For our estimates, we choose [latex]x=0.00001[\/latex] and [latex]x=-0.00001[\/latex] to obtain the estimate<\/p>\r\n\r\n<div id=\"fs-id1169737725568\" class=\"equation unnumbered\">[latex]\\frac{b^{-0.00001}-1}{-0.00001}&lt;B^{\\prime}(0)&lt;\\frac{b^{0.00001}-1}{0.00001}[\/latex].<\/div>\r\n<p id=\"fs-id1169737952624\">See the following table.<\/p>\r\n\r\n<table id=\"fs-id1169738019199\" summary=\"This table has six rows and four columns. The first row is a header row and it labels each column. The first column header is b, the second column header is (b\u22120.00001 \u2212 1)\/\u22120.00001 &lt; B\u2019(0) &lt; (b0.00001 \u2212 1)\/0.00001, the third column header is b, and the fourth column header is (b\u22120.00001 \u2212 1)\/\u22120.00001 &lt; B\u2019(0) &lt; (b0.00001 \u2212 1)\/0.00001. Under the first column are the values 2, 2.7, 2.71, 2.718, and 2.7182. Under the second column are the values 0.693145&lt;B\u2019(0)&lt;0.69315, 0.993247&lt;B\u2019(0)&lt; 0.993257, 0.996944&lt;B\u2019(0)&lt;0.996954, 0.999891&lt;B\u2019(0)&lt; 0.999901, and 0.999965&lt;B\u2019(0)&lt;0.999975. Under the third column are the values 2.7183, 2.719, 2.72, 2.8, and 3. Under the fourth column are the values 1.000002&lt;B\u2019(0)&lt; 1.000012, 1.000259&lt;B\u2019(0)&lt; 1.000269, 1.000627&lt;B\u2019(0)&lt;1.000637, 1.029614&lt;B\u2019(0)&lt;1.029625, and 1.098606&lt;B\u2019(00&lt;1.098618.\"><caption>Estimating a Value of [latex]e[\/latex]<\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>[latex]b[\/latex]<\/th>\r\n<th>[latex]\\frac{b^{-0.00001}-1}{-0.00001}&lt;B^{\\prime}(0)&lt;\\frac{b^{0.00001}-1}{0.00001}[\/latex]<\/th>\r\n<th>[latex]b[\/latex]<\/th>\r\n<th>[latex]\\frac{b^{-0.00001}-1}{-0.00001}&lt;B^{\\prime}(0)&lt;\\frac{b^{0.00001}-1}{0.00001}[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>2<\/td>\r\n<td>[latex]0.693145&lt;B^{\\prime}(0)&lt;0.69315[\/latex]<\/td>\r\n<td>2.7183<\/td>\r\n<td>[latex]1.000002&lt;B^{\\prime}(0)&lt;1.000012[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2.7<\/td>\r\n<td>[latex]0.993247&lt;B^{\\prime}(0)&lt;0.993257[\/latex]<\/td>\r\n<td>2.719<\/td>\r\n<td>[latex]1.000259&lt;B^{\\prime}(0)&lt;1.000269[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2.71<\/td>\r\n<td>[latex]0.996944&lt;B^{\\prime}(0)&lt;0.996954[\/latex]<\/td>\r\n<td>2.72<\/td>\r\n<td>[latex]1.000627&lt;B^{\\prime}(0)&lt;1.000637[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2.718<\/td>\r\n<td>[latex]0.999891&lt;B^{\\prime}(0)&lt;0.999901[\/latex]<\/td>\r\n<td>2.8<\/td>\r\n<td>[latex]1.029614&lt;B^{\\prime}(0)&lt;1.029625[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>2.7182<\/td>\r\n<td>[latex]0.999965&lt;B^{\\prime}(0)&lt;0.999975[\/latex]<\/td>\r\n<td>3<\/td>\r\n<td>[latex]1.098606&lt;B^{\\prime}(0)&lt;1.098618[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1169738213815\">The evidence from the table suggests that [latex]2.7182&lt;e&lt;2.7183[\/latex].<\/p>\r\n<p id=\"fs-id1169737978597\">The graph of [latex]E(x)=e^x[\/latex] together with the line [latex]y=x+1[\/latex] are shown in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_03_09_002\">(Figure)<\/a>. This line is tangent to the graph of [latex]E(x)=e^x[\/latex] at [latex]x=0[\/latex].<\/p>\r\n\r\n<div id=\"CNX_Calc_Figure_03_09_002\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205533\/CNX_Calc_Figure_03_09_002.jpg\" alt=\"Graph of the function ex along with its tangent at (0, 1), x + 1.\" width=\"487\" height=\"248\" \/> <strong>Figure 2.<\/strong> The tangent line to [latex]E(x)=e^x[\/latex] at [latex]x=0[\/latex] has slope 1.[\/caption]<\/div>\r\n<div class=\"wp-caption-text\"><\/div>\r\n<p id=\"fs-id1169738198749\">Now that we have laid out our basic assumptions, we begin our investigation by exploring the derivative of [latex]B(x)=b^x, \\, b&gt;0[\/latex]. Recall that we have assumed that [latex]B^{\\prime}(0)[\/latex] exists. By applying the limit definition to the derivative we conclude that<\/p>\r\n\r\n<div id=\"fs-id1169738045860\" class=\"equation\">[latex]B^{\\prime}(0)=\\underset{h\\to 0}{\\lim}\\frac{b^{0+h}-b^0}{h}=\\underset{h\\to 0}{\\lim}\\frac{b^h-1}{h}[\/latex].<\/div>\r\n<p id=\"fs-id1169738187820\">Turning to [latex]B^{\\prime}(x)[\/latex], we obtain the following.<\/p>\r\n\r\n<div id=\"fs-id1169737954073\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} B^{\\prime}(x) &amp; =\\underset{h\\to 0}{\\lim}\\frac{b^{x+h}-b^x}{h} &amp; &amp; &amp; \\text{Apply the limit definition of the derivative.} \\\\ &amp; =\\underset{h\\to 0}{\\lim}\\frac{b^xb^h-b^x}{h} &amp; &amp; &amp; \\text{Note that} \\, b^{x+h}=b^x b^h. \\\\ &amp; =\\underset{h\\to 0}{\\lim}\\frac{b^x(b^h-1)}{h} &amp; &amp; &amp; \\text{Factor out} \\, b^x. \\\\ &amp; =b^x\\underset{h\\to 0}{\\lim}\\frac{b^h-1}{h} &amp; &amp; &amp; \\text{Apply a property of limits.} \\\\ &amp; =b^x B^{\\prime}(0) &amp; &amp; &amp; \\text{Use} \\, B^{\\prime}(0)=\\underset{h\\to 0}{\\lim}\\frac{b^{0+h}-b^0}{h}=\\underset{h\\to 0}{\\lim}\\frac{b^h-1}{h}. \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1169738221160\">We see that on the basis of the assumption that [latex]B(x)=b^x[\/latex] is differentiable at [latex]0, \\, B(x)[\/latex] is not only differentiable everywhere, but its derivative is<\/p>\r\n\r\n<div id=\"fs-id1169738221385\" class=\"equation\">[latex]B^{\\prime}(x)=b^x B^{\\prime}(0)[\/latex].<\/div>\r\n<p id=\"fs-id1169738183261\">For [latex]E(x)=e^x, \\, E^{\\prime}(0)=1[\/latex]. Thus, we have [latex]E^{\\prime}(x)=e^x[\/latex]. (The value of [latex]B^{\\prime}(0)[\/latex] for an arbitrary function of the form [latex]B(x)=b^x, \\, b&gt;0[\/latex], will be derived later.)<\/p>\r\n\r\n<div id=\"fs-id1169738226753\" class=\"textbox key-takeaways theorem\">\r\n<h3>Derivative of the Natural Exponential Function<\/h3>\r\n<p id=\"fs-id1169738220224\">Let [latex]E(x)=e^x[\/latex] be the natural exponential function. Then<\/p>\r\n\r\n<div id=\"fs-id1169737928243\" class=\"equation unnumbered\">[latex]E^{\\prime}(x)=e^x[\/latex].<\/div>\r\n<p id=\"fs-id1169737142141\">In general,<\/p>\r\n\r\n<div id=\"fs-id1169738124964\" class=\"equation unnumbered\">[latex]\\frac{d}{dx}(e^{g(x)})=e^{g(x)} g^{\\prime}(x)[\/latex].<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738223456\" class=\"textbox examples\">\r\n<h3>Derivative of an Exponential Function<\/h3>\r\n<div id=\"fs-id1169738223458\" class=\"exercise\">\r\n<div id=\"fs-id1169738223460\" class=\"textbox\">\r\n<p id=\"fs-id1169738187154\">Find the derivative of [latex]f(x)=e^{\\tan (2x)}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169737140844\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737140844\"]\r\n<p id=\"fs-id1169737140844\">Using the derivative formula and the chain rule,<\/p>\r\n\r\n<div id=\"fs-id1169738048872\" class=\"equation unnumbered\">[latex]\\begin{array}{ll} f^{\\prime}(x) &amp; =e^{\\tan (2x)}\\frac{d}{dx}(\\tan (2x)) \\\\ &amp; = e^{\\tan (2x)} \\sec^2 (2x) \\cdot 2. \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737140879\" class=\"textbox examples\">\r\n<h3>Combining Differentiation Rules<\/h3>\r\n<div id=\"fs-id1169737140881\" class=\"exercise\">\r\n<div id=\"fs-id1169737766541\" class=\"textbox\">\r\n<p id=\"fs-id1169737766547\">Find the derivative of [latex]y=\\frac{e^{x^2}}{x}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169737928258\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737928258\"]\r\n<p id=\"fs-id1169737928258\">Use the derivative of the natural exponential function, the quotient rule, and the chain rule.<\/p>\r\n\r\n<div id=\"fs-id1169737928262\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} y^{\\prime} &amp; =\\large \\frac{(e^{x^2} \\cdot 2x) \\cdot x - 1 \\cdot e^{x^2}}{x^2} &amp; &amp; &amp; \\text{Apply the quotient rule.} \\\\ &amp; = \\large \\frac{e^{x^2}(2x^2-1)}{x^2} &amp; &amp; &amp; \\text{Simplify.} \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738152550\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1169738152553\" class=\"exercise\">\r\n<div id=\"fs-id1169737923849\" class=\"textbox\">\r\n<p id=\"fs-id1169737923851\">Find the derivative of [latex]h(x)=xe^{2x}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169737948362\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737948362\"]\r\n<p id=\"fs-id1169737948362\">[latex]h^{\\prime}(x)=e^{2x}+2xe^{2x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1169738222020\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169738215078\">Don\u2019t forget to use the product rule.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737949367\" class=\"textbox examples\">\r\n<h3>Applying the Natural Exponential Function<\/h3>\r\n<div id=\"fs-id1169737949369\" class=\"exercise\">\r\n<div id=\"fs-id1169737151938\" class=\"textbox\">\r\n<p id=\"fs-id1169737151943\">A colony of mosquitoes has an initial population of 1000. After [latex]t[\/latex] days, the population is given by [latex]A(t)=1000e^{0.3t}[\/latex]. Show that the ratio of the rate of change of the population, [latex]A^{\\prime}(t)[\/latex], to the population size, [latex]A(t)[\/latex] is constant.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738212487\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738212487\"]\r\n<p id=\"fs-id1169738212487\">First find [latex]A^{\\prime}(t)[\/latex]. By using the chain rule, we have [latex]A^{\\prime}(t)=300e^{0.3t}[\/latex]. Thus, the ratio of the rate of change of the population to the population size is given by<\/p>\r\n\r\n<div id=\"fs-id1169738227689\" class=\"equation unnumbered\">[latex]\\large \\frac{A^{\\prime}(t)}{A(t)} \\normalsize = \\large \\frac{300e^{0.3t}}{1000e^{0.3t}}=0.3[\/latex].<\/div>\r\n<p id=\"fs-id1169737145031\">The ratio of the rate of change of the population to the population size is the constant 0.3.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738186819\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1169738186822\" class=\"exercise\">\r\n<div id=\"fs-id1169738186824\" class=\"textbox\">\r\n<p id=\"fs-id1169738215955\">If [latex]A(t)=1000e^{0.3t}[\/latex] describes the mosquito population after [latex]t[\/latex] days, as in the preceding example, what is the rate of change of [latex]A(t)[\/latex] after 4 days?<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169737934408\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737934408\"]\r\n<p id=\"fs-id1169737934408\">996<\/p>\r\n\r\n<\/div>\r\n<div class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169738225603\">Find [latex]A^{\\prime}(4)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738221999\" class=\"bc-section section\">\r\n<h1>Derivative of the Logarithmic Function<\/h1>\r\n<p id=\"fs-id1169738222225\">Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function.<\/p>\r\n\r\n<div id=\"fs-id1169737927590\" class=\"textbox key-takeaways theorem\">\r\n<h3>The Derivative of the Natural Logarithmic Function<\/h3>\r\n<p id=\"fs-id1169737911417\">If [latex]x&gt;0[\/latex] and [latex]y=\\ln x[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1169738223534\" class=\"equation\">[latex]\\frac{dy}{dx}=\\frac{1}{x}[\/latex].<\/div>\r\n<p id=\"fs-id1169737765282\">More generally, let [latex]g(x)[\/latex] be a differentiable function. For all values of [latex]x[\/latex] for which [latex]g^{\\prime}(x)&gt;0[\/latex], the derivative of [latex]h(x)=\\ln(g(x))[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1169737919348\" class=\"equation\">[latex]h^{\\prime}(x)=\\frac{1}{g(x)} g^{\\prime}(x)[\/latex].<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738093053\" class=\"bc-section section\">\r\n<h2>Proof<\/h2>\r\n<p id=\"fs-id1169738186955\">If [latex]x&gt;0[\/latex] and [latex]y=\\ln x[\/latex], then [latex]e^y=x[\/latex]. Differentiating both sides of this equation results in the equation<\/p>\r\n\r\n<div id=\"fs-id1169738212688\" class=\"equation unnumbered\">[latex]e^y\\frac{dy}{dx}=1[\/latex].<\/div>\r\n<p id=\"fs-id1169738216993\">Solving for [latex]\\frac{dy}{dx}[\/latex] yields<\/p>\r\n\r\n<div class=\"equation unnumbered\">[latex]\\frac{dy}{dx}=\\frac{1}{e^y}[\/latex].<\/div>\r\n<p id=\"fs-id1169738219487\">Finally, we substitute [latex]x=e^y[\/latex] to obtain<\/p>\r\n\r\n<div id=\"fs-id1169737145231\" class=\"equation unnumbered\">[latex]\\frac{dy}{dx}=\\frac{1}{x}[\/latex].<\/div>\r\n<p id=\"fs-id1169738221909\">We may also derive this result by applying the inverse function theorem, as follows. Since [latex]y=g(x)=\\ln x[\/latex] is the inverse of [latex]f(x)=e^x[\/latex], by applying the inverse function theorem we have<\/p>\r\n\r\n<div id=\"fs-id1169738070944\" class=\"equation unnumbered\">[latex]\\frac{dy}{dx}=\\frac{1}{f^{\\prime}(g(x))}=\\frac{1}{e^{\\ln x}}=\\frac{1}{x}[\/latex].<\/div>\r\n<p id=\"fs-id1169737951681\">Using this result and applying the chain rule to [latex]h(x)=\\ln(g(x))[\/latex] yields<\/p>\r\n\r\n<div id=\"fs-id1169738106180\" class=\"equation unnumbered\">[latex]h^{\\prime}(x)=\\frac{1}{g(x)} g^{\\prime}(x). _\\blacksquare[\/latex]<\/div>\r\nThe graph of [latex]y=\\ln x[\/latex] and its derivative [latex]\\frac{dy}{dx}=\\frac{1}{x}[\/latex] are shown in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_03_09_003\">(Figure)<\/a>.\r\n<div id=\"CNX_Calc_Figure_03_09_003\" class=\"wp-caption aligncenter\">[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205536\/CNX_Calc_Figure_03_09_003.jpg\" alt=\"Graph of the function ln x along with its derivative 1\/x. The function ln x is increasing on (0, + \u221e). Its derivative is decreasing but greater than 0 on (0, + \u221e).\" width=\"487\" height=\"209\" \/> <strong> Figure 3.<\/strong>\u00a0The function [latex]y=\\ln x[\/latex] is increasing on [latex](0,+\\infty)[\/latex]. Its derivative [latex]y^{\\prime} =\\frac{1}{x}[\/latex] is greater than zero on [latex](0,+\\infty)[\/latex].[\/caption]<\/div>\r\n<div id=\"fs-id1169738211098\" class=\"textbox examples\">\r\n<h3>Taking a Derivative of a Natural Logarithm<\/h3>\r\n<div id=\"fs-id1169738228370\" class=\"exercise\">\r\n<div id=\"fs-id1169738228372\" class=\"textbox\">\r\n<p id=\"fs-id1169738228377\">Find the derivative of [latex]f(x)=\\ln(x^3+3x-4)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738221312\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738221312\"]\r\n<p id=\"fs-id1169738221312\">Use <a class=\"autogenerated-content\" href=\"#fs-id1169737919348\">(Figure)<\/a> directly.<\/p>\r\n\r\n<div id=\"fs-id1169738220266\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} f^{\\prime}(x) &amp; =\\frac{1}{x^3+3x-4} \\cdot (3x^2+3) &amp; &amp; &amp; \\text{Use} \\, g(x)=x^3+3x-4 \\, \\text{in} \\, h^{\\prime}(x)=\\frac{1}{g(x)} g^{\\prime}(x). \\\\ &amp; =\\frac{3x^2+3}{x^3+3x-4} &amp; &amp; &amp; \\text{Rewrite.} \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738244488\" class=\"textbox examples\">\r\n<h3>Using Properties of Logarithms in a Derivative<\/h3>\r\n<div id=\"fs-id1169738106136\" class=\"exercise\">\r\n<div id=\"fs-id1169738106138\" class=\"textbox\">\r\n<p id=\"fs-id1169738106144\">Find the derivative of [latex]f(x)=\\ln(\\frac{x^2 \\sin x}{2x+1})[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738219674\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738219674\"]\r\n<p id=\"fs-id1169738219674\">At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.<\/p>\r\n\r\n<div id=\"fs-id1169738219679\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} f(x) &amp; = \\ln(\\frac{x^2 \\sin x}{2x+1})=2\\ln x+\\ln(\\sin x)-\\ln(2x+1) &amp; &amp; &amp; \\text{Apply properties of logarithms.} \\\\ f^{\\prime}(x) &amp; = \\frac{2}{x} + \\frac{\\cos x}{\\sin x} -\\frac{2}{2x+1} &amp; &amp; &amp; \\text{Apply sum rule and} \\, h^{\\prime}(x)=\\frac{1}{g(x)} g^{\\prime}(x). \\\\ &amp; = \\frac{2}{x} + \\cot x - \\frac{2}{2x+1} &amp; &amp; &amp; \\text{Simplify using the quotient identity for cotangent.} \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738219654\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1169738219657\" class=\"exercise\">\r\n<div id=\"fs-id1169738219659\" class=\"textbox\">\r\n<p id=\"fs-id1169738219661\">Differentiate: [latex]f(x)=\\ln (3x+2)^5[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738192196\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738192196\"]\r\n<p id=\"fs-id1169738192196\">[latex]f^{\\prime}(x)=\\frac{15}{3x+2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1169738073195\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169738073202\">Use a property of logarithms to simplify before taking the derivative.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169738076342\">Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of [latex]y=\\log_b x[\/latex] and [latex]y=b^x[\/latex] for [latex]b&gt;0, \\, b\\ne 1[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169738238181\" class=\"textbox key-takeaways theorem\">\r\n<h3>Derivatives of General Exponential and Logarithmic Functions<\/h3>\r\n<p id=\"fs-id1169738186170\">Let [latex]b&gt;0, \\, b\\ne 1[\/latex], and let [latex]g(x)[\/latex] be a differentiable function.<\/p>\r\n\r\n<ol id=\"fs-id1169737998025\">\r\n \t<li>If [latex]y=\\log_b x[\/latex], then\r\n<div id=\"fs-id1169738105090\" class=\"equation\">[latex]\\frac{dy}{dx}=\\frac{1}{x \\ln b}[\/latex].<\/div>\r\nMore generally, if [latex]h(x)=\\log_b (g(x))[\/latex], then for all values of [latex]x[\/latex] for which [latex]g(x)&gt;0[\/latex],\r\n<div id=\"fs-id1169738186308\" class=\"equation\">[latex]h^{\\prime}(x)=\\frac{g^{\\prime}(x)}{g(x) \\ln b}[\/latex].\r\n\r\n<\/div><\/li>\r\n \t<li>If [latex]y=b^x[\/latex], then\r\n<div id=\"fs-id1169738224034\" class=\"equation\">[latex]\\frac{dy}{dx}=b^x \\ln b[\/latex].<\/div>\r\nMore generally, if [latex]h(x)=b^{g(x)}[\/latex], then\r\n<div id=\"fs-id1169738045159\" class=\"equation\">[latex]h^{\\prime}(x)=b^{g(x)} g^{\\prime}(x) \\ln b[\/latex].<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bc-section section\">\r\n<h2>Proof<\/h2>\r\n<p id=\"fs-id1169738184804\">If [latex]y=\\log_b x[\/latex], then [latex]b^y=x[\/latex]. It follows that [latex]\\ln(b^y)=\\ln x[\/latex]. Thus [latex]y \\ln b = \\ln x[\/latex]. Solving for [latex]y[\/latex], we have [latex]y=\\frac{\\ln x}{\\ln b}[\/latex]. Differentiating and keeping in mind that [latex]\\ln b[\/latex] is a constant, we see that<\/p>\r\n\r\n<div id=\"fs-id1169738211798\" class=\"equation unnumbered\">[latex]\\frac{dy}{dx}=\\frac{1}{x \\ln b}[\/latex].<\/div>\r\n<p id=\"fs-id1169738211864\">The derivative in <a class=\"autogenerated-content\" href=\"#fs-id1169738186308\">(Figure)<\/a> now follows from the chain rule.<\/p>\r\n<p id=\"fs-id1169737954089\">If [latex]y=b^x[\/latex], then [latex]\\ln y=x \\ln b[\/latex]. Using implicit differentiation, again keeping in mind that [latex]\\ln b[\/latex] is constant, it follows that [latex]\\frac{1}{y}\\frac{dy}{dx}=\\text{ln}b.[\/latex] Solving for [latex]\\frac{dy}{dx}[\/latex] and substituting [latex]y=b^x[\/latex], we see that<\/p>\r\n\r\n<div id=\"fs-id1169737934328\" class=\"equation unnumbered\">[latex]\\frac{dy}{dx}=y \\ln b=b^x \\ln b[\/latex].<\/div>\r\n<p id=\"fs-id1169737145057\">The more general derivative (<a class=\"autogenerated-content\" href=\"#fs-id1169738045159\">(Figure)<\/a>) follows from the chain rule. [latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1169737145066\" class=\"textbox examples\">\r\n<h3>Applying Derivative Formulas<\/h3>\r\n<div class=\"exercise\">\r\n<div id=\"fs-id1169738185254\" class=\"textbox\">\r\n<p id=\"fs-id1169738185259\">Find the derivative of [latex]h(x)=\\large \\frac{3^x}{3^x+2}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169737700313\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737700313\"]\r\n<p id=\"fs-id1169737700313\">Use the quotient rule and <a class=\"autogenerated-content\" href=\"#fs-id1169738238181\">(Figure)<\/a>.<\/p>\r\n\r\n<div id=\"fs-id1169737700320\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} h^{\\prime}(x) &amp; = \\large \\frac{3^x \\ln 3(3^x+2)-3^x \\ln 3(3^x)}{(3^x+2)^2} &amp; &amp; &amp; \\text{Apply the quotient rule.} \\\\ &amp; = \\large \\frac{2 \\cdot 3^x \\ln 3}{(3^x+2)^2} &amp; &amp; &amp; \\text{Simplify.} \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738219393\" class=\"textbox examples\">\r\n<h3>Finding the Slope of a Tangent Line<\/h3>\r\n<div id=\"fs-id1169738219395\" class=\"exercise\">\r\n<div id=\"fs-id1169738219397\" class=\"textbox\">\r\n<p id=\"fs-id1169738219403\">Find the slope of the line tangent to the graph of [latex]y=\\log_2 (3x+1)[\/latex] at [latex]x=1[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738045067\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738045067\"]\r\n<p id=\"fs-id1169738045067\">To find the slope, we must evaluate [latex]\\frac{dy}{dx}[\/latex] at [latex]x=1[\/latex]. Using <a class=\"autogenerated-content\" href=\"#fs-id1169738186308\">(Figure)<\/a>, we see that<\/p>\r\n\r\n<div id=\"fs-id1169738197861\" class=\"equation unnumbered\">[latex]\\frac{dy}{dx}=\\frac{3}{\\ln 2(3x+1)}[\/latex].<\/div>\r\n<p id=\"fs-id1169738186987\">By evaluating the derivative at [latex]x=1[\/latex], we see that the tangent line has slope<\/p>\r\n\r\n<div id=\"fs-id1169738187002\" class=\"equation unnumbered\">[latex]\\frac{dy}{dx}|_{x=1} =\\frac{3}{4 \\ln 2}=\\frac{3}{\\ln 16}[\/latex].[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738240282\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1169738240285\" class=\"exercise\">\r\n<div id=\"fs-id1169738240288\" class=\"textbox\">\r\n<p id=\"fs-id1169738240290\">Find the slope for the line tangent to [latex]y=3^x[\/latex] at [latex]x=2[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169737934288\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737934288\"]\r\n<p id=\"fs-id1169737934288\">[latex]9 \\ln (3)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1169737934307\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169737934313\">Evaluate the derivative at [latex]x=2[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737933453\" class=\"bc-section section\">\r\n<h1>Logarithmic Differentiation<\/h1>\r\n<p id=\"fs-id1169737933458\">At this point, we can take derivatives of functions of the form [latex]y=(g(x))^n[\/latex] for certain values of [latex]n[\/latex], as well as functions of the form [latex]y=b^{g(x)}[\/latex], where [latex]b&gt;0[\/latex] and [latex]b\\ne 1[\/latex]. Unfortunately, we still do not know the derivatives of functions such as [latex]y=x^x[\/latex] or [latex]y=x^{\\pi}[\/latex]. These functions require a technique called<strong> logarithmic differentiation<\/strong>, which allows us to differentiate any function of the form [latex]h(x)=g(x)^{f(x)}[\/latex]. It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of [latex]y=\\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x}[\/latex]. We outline this technique in the following problem-solving strategy.<\/p>\r\n\r\n<div id=\"fs-id1169738197957\" class=\"textbox key-takeaways problem-solving\">\r\n<h3>Problem-Solving Strategy: Using Logarithmic Differentiation<\/h3>\r\n<ol id=\"fs-id1169738197963\">\r\n \t<li>To differentiate [latex]y=h(x)[\/latex] using logarithmic differentiation, take the natural logarithm of both sides of the equation to obtain [latex]\\ln y=\\ln (h(x))[\/latex].<\/li>\r\n \t<li>Use properties of logarithms to expand [latex]\\ln (h(x))[\/latex] as much as possible.<\/li>\r\n \t<li>Differentiate both sides of the equation. On the left we will have [latex]\\frac{1}{y}\\frac{dy}{dx}[\/latex].<\/li>\r\n \t<li>Multiply both sides of the equation by [latex]y[\/latex] to solve for [latex]\\frac{dy}{dx}[\/latex].<\/li>\r\n \t<li>Replace [latex]y[\/latex] by [latex]h(x)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1169738238112\" class=\"textbox examples\">\r\n<h3>Using Logarithmic Differentiation<\/h3>\r\n<div id=\"fs-id1169738238115\" class=\"exercise\">\r\n<div id=\"fs-id1169738238117\" class=\"textbox\">\r\n<p id=\"fs-id1169738211866\">Find the derivative of [latex]y=(2x^4+1)^{\\tan x}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738211908\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738211908\"]\r\n<p id=\"fs-id1169738211908\">Use logarithmic differentiation to find this derivative.<\/p>\r\n\r\n<div id=\"fs-id1169738211912\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} \\ln y &amp; = \\ln (2x^4+1)^{\\tan x} &amp; &amp; &amp; \\text{Step 1. Take the natural logarithm of both sides.} \\\\ \\ln y &amp; = \\tan x \\ln (2x^4+1) &amp; &amp; &amp; \\text{Step 2. Expand using properties of logarithms.} \\\\ \\frac{1}{y}\\frac{dy}{dx} &amp; = \\sec^2 x \\ln (2x^4+1)+\\frac{8x^3}{2x^4+1} \\cdot \\tan x &amp; &amp; &amp; \\begin{array}{l}\\text{Step 3. Differentiate both sides. Use the} \\\\ \\text{product rule on the right.} \\end{array} \\\\ \\frac{dy}{dx} &amp; =y \\cdot (\\sec^2 x \\ln (2x^4+1)+\\frac{8x^3}{2x^4+1} \\cdot \\tan x) &amp; &amp; &amp; \\text{Step 4. Multiply by} \\, y \\, \\text{on both sides.} \\\\ \\frac{dy}{dx} &amp; = (2x^4+1)^{\\tan x}(\\sec^2 x \\ln (2x^4+1)+\\frac{8x^3}{2x^4+1} \\cdot \\tan x) &amp; &amp; &amp; \\text{Step 5. Substitute} \\, y=(2x^4+1)^{\\tan x}.\\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738068346\" class=\"textbox examples\">\r\n<h3>Using Logarithmic Differentiation<\/h3>\r\n<div id=\"fs-id1169738068348\" class=\"exercise\">\r\n<div id=\"fs-id1169738068350\" class=\"textbox\">\r\n<p id=\"fs-id1169738068356\">Find the derivative of [latex]y=\\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1169738201884\" class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738201884\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738201884\"]This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter.\r\n<div id=\"fs-id1169738201891\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} \\ln y &amp; = \\ln \\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x} &amp; &amp; &amp; \\text{Step 1. Take the natural logarithm of both sides.} \\\\ \\ln y &amp; = \\ln x+\\frac{1}{2} \\ln (2x+1)-x \\ln e-3 \\ln \\sin x &amp; &amp; &amp; \\text{Step 2. Expand using properties of logarithms.} \\\\ \\frac{1}{y}\\frac{dy}{dx} &amp; = \\frac{1}{x}+\\frac{1}{2x+1}-1-3\\big(\\frac{\\cos x}{\\sin x}\\big) &amp; &amp; &amp; \\text{Step 3. Differentiate both sides.} \\\\ \\frac{dy}{dx} &amp; = y (\\frac{1}{x}+\\frac{1}{2x+1}-1-3 \\cot x) &amp; &amp; &amp; \\text{Step 4. Multiply by} \\, y \\, \\text{on both sides and simplify.} \\\\ \\frac{dy}{dx} &amp; = \\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x} \\normalsize (\\frac{1}{x}+\\frac{1}{2x+1}-1-3 \\cot x) &amp; &amp; &amp; \\text{Step 5. Substitute} \\, y=\\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x}. \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738228453\" class=\"textbox examples\">\r\n<h3>Extending the Power Rule<\/h3>\r\n<div id=\"fs-id1169738228455\" class=\"exercise\">\r\n<div id=\"fs-id1169738228457\" class=\"textbox\">\r\n<p id=\"fs-id1169738228463\">Find the derivative of [latex]y=x^r[\/latex] where [latex]r[\/latex] is an arbitrary real number.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738228486\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738228486\"]\r\n<p id=\"fs-id1169738228486\">The process is the same as in <a class=\"autogenerated-content\" href=\"#fs-id1169738068346\">(Figure)<\/a>, though with fewer complications.<\/p>\r\n\r\n<div id=\"fs-id1169738228494\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} \\ln y &amp; = \\ln x^r &amp; &amp; &amp; \\text{Step 1. Take the natural logarithm of both sides.} \\\\ \\ln y &amp; = r \\ln x &amp; &amp; &amp; \\text{Step 2. Expand using properties of logarithms.} \\\\ \\frac{1}{y}\\frac{dy}{dx} &amp; = r \\frac{1}{x} &amp; &amp; &amp; \\text{Step 3. Differentiate both sides.} \\\\ \\frac{dy}{dx} &amp; = y \\frac{r}{x} &amp; &amp; &amp; \\text{Step 4. Multiply by} \\, y \\, \\text{on both sides.} \\\\ \\frac{dy}{dx} &amp; = x^r \\frac{r}{x} &amp; &amp; &amp; \\text{Step 5. Substitute} \\, y=x^r. \\\\ \\frac{dy}{dx} &amp; = rx^{r-1} &amp; &amp; &amp; \\text{Simplify.} \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737933509\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1169737933513\" class=\"exercise\">\r\n<div id=\"fs-id1169737933515\" class=\"textbox\">\r\n<p id=\"fs-id1169737933517\">Use logarithmic differentiation to find the derivative of [latex]y=x^x[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169737933537\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737933537\"]\r\n<p id=\"fs-id1169737933537\">[latex]\\frac{dy}{dx}=x^x(1+\\ln x)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169737933587\">Follow the problem solving strategy.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737933594\" class=\"textbox exercises checkpoint\">\r\n<div id=\"fs-id1169737933598\" class=\"exercise\">\r\n<div id=\"fs-id1169737933600\" class=\"textbox\">\r\n<p id=\"fs-id1169737933602\">Find the derivative of [latex]y=(\\tan x)^{\\pi}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738233543\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738233543\"]\r\n<p id=\"fs-id1169738233543\">[latex]y^{\\prime}=\\pi (\\tan x)^{\\pi -1} \\sec^2 x[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1169738233590\" class=\"commentary\">\r\n<h4>Hint<\/h4>\r\n<p id=\"fs-id1169738233597\">Use the result from <a class=\"autogenerated-content\" href=\"#fs-id1169738228453\">(Figure)<\/a>.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738233608\" class=\"textbox key-takeaways\">\r\n<h3>Key Concepts<\/h3>\r\n<ul id=\"fs-id1169738233616\">\r\n \t<li>On the basis of the assumption that the exponential function [latex]y=b^x, \\, b&gt;0[\/latex] is continuous everywhere and differentiable at 0, this function is differentiable everywhere and there is a formula for its derivative.<\/li>\r\n \t<li>We can use a formula to find the derivative of [latex]y=\\ln x[\/latex], and the relationship [latex]\\log_b x=\\frac{\\ln x}{\\ln b}[\/latex] allows us to extend our differentiation formulas to include logarithms with arbitrary bases.<\/li>\r\n \t<li>Logarithmic differentiation allows us to differentiate functions of the form [latex]y=g(x)^{f(x)}[\/latex] or very complex functions by taking the natural logarithm of both sides and exploiting the properties of logarithms before differentiating.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1169738234614\" class=\"key-equations\">\r\n<h1>Key Equations<\/h1>\r\n<ul id=\"fs-id1169738234621\">\r\n \t<li><strong>Derivative of the natural exponential function<\/strong>\r\n[latex]\\frac{d}{dx}(e^{g(x)})=e^{g(x)} g^{\\prime}(x)[\/latex]<\/li>\r\n \t<li><strong>Derivative of the natural logarithmic function<\/strong>\r\n[latex]\\frac{d}{dx}(\\ln (g(x)))=\\frac{1}{g(x)} g^{\\prime}(x)[\/latex]<\/li>\r\n \t<li><strong>Derivative of the general exponential function<\/strong>\r\n[latex]\\frac{d}{dx}(b^{g(x)})=b^{g(x)} g^{\\prime}(x) \\ln b[\/latex]<\/li>\r\n \t<li><strong>Derivative of the general logarithmic function<\/strong>\r\n[latex]\\frac{d}{dx}(\\log_b (g(x)))=\\frac{g^{\\prime}(x)}{g(x) \\ln b}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1169738235113\" class=\"textbox exercises\">\r\n<p id=\"fs-id1169738235117\">For the following exercises, find [latex]f^{\\prime}(x)[\/latex] for each function.<\/p>\r\n\r\n<div id=\"fs-id1169738235136\" class=\"exercise\">\r\n<div id=\"fs-id1169738235139\" class=\"textbox\">\r\n<p id=\"fs-id1169738235141\"><strong>1.\u00a0<\/strong>[latex]f(x)=x^2 e^x[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738235171\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738235171\"]\r\n<p id=\"fs-id1169738235171\">[latex]f^{\\prime}(x) = 2xe^x+x^2 e^x[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738235199\" class=\"exercise\">\r\n<div id=\"fs-id1169738235202\" class=\"textbox\">\r\n<p id=\"fs-id1169738235204\"><strong>2.\u00a0<\/strong>[latex]f(x)=\\frac{e^{\u2212x}}{x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738085451\" class=\"exercise\">\r\n<div id=\"fs-id1169738085453\" class=\"textbox\">\r\n<p id=\"fs-id1169738085455\"><strong>3.\u00a0<\/strong>[latex]f(x)=e^{x^3 \\ln x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738085491\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738085491\"]\r\n<p id=\"fs-id1169738085491\">[latex]f^{\\prime}(x) = e^{x^3 \\ln x}(3x^2 \\ln x+x^2)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738085543\" class=\"exercise\">\r\n<div id=\"fs-id1169738085545\" class=\"textbox\">\r\n<p id=\"fs-id1169738085548\"><strong>4.\u00a0<\/strong>[latex]f(x)=\\sqrt{e^{2x}+2x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738240382\" class=\"exercise\">\r\n<div id=\"fs-id1169738240384\" class=\"textbox\">\r\n<p id=\"fs-id1169738240386\"><strong>5.\u00a0<\/strong>[latex]f(x)=\\large \\frac{e^x-e^{\u2212x}}{e^x+e^{\u2212x}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738240440\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738240440\"]\r\n<p id=\"fs-id1169738240440\">[latex]f^{\\prime}(x) = \\large \\frac{4}{(e^x+e^{\u2212x})^2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738240478\" class=\"exercise\">\r\n<div id=\"fs-id1169738240480\" class=\"textbox\">\r\n<p id=\"fs-id1169738240482\"><strong>6.\u00a0<\/strong>[latex]f(x)=\\frac{10^x}{\\ln 10}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738071335\" class=\"exercise\">\r\n<div id=\"fs-id1169738071337\" class=\"textbox\">\r\n<p id=\"fs-id1169738071339\"><strong>7.\u00a0<\/strong>[latex]f(x)=2^{4x}+4x^2[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738071376\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738071376\"]\r\n<p id=\"fs-id1169738071376\">[latex]f^{\\prime}(x) = 2^{4x+2} \\cdot \\ln 2+8x[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738071412\" class=\"exercise\">\r\n<div id=\"fs-id1169738071414\" class=\"textbox\">\r\n<p id=\"fs-id1169738071416\"><strong>8.\u00a0<\/strong>[latex]f(x)=3^{\\sin 3x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738039846\" class=\"exercise\">\r\n<div id=\"fs-id1169738039848\" class=\"textbox\">\r\n<p id=\"fs-id1169738039850\"><strong>9.\u00a0<\/strong>[latex]f(x)=x^{\\pi} \\cdot \\pi^x[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738039882\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738039882\"]\r\n<p id=\"fs-id1169738039882\">[latex]f^{\\prime}(x) = \\pi x^{\\pi -1} \\cdot \\pi^x + x^{\\pi} \\cdot \\pi^x \\ln \\pi [\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738039931\" class=\"exercise\">\r\n<div id=\"fs-id1169738039933\" class=\"textbox\">\r\n<p id=\"fs-id1169738039935\"><strong>10.\u00a0<\/strong>[latex]f(x)=\\ln(4x^3+x)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737904496\" class=\"exercise\">\r\n<div id=\"fs-id1169737904498\" class=\"textbox\">\r\n<p id=\"fs-id1169737904500\"><strong>11.\u00a0<\/strong>[latex]f(x)=\\ln \\sqrt{5x-7}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169737904532\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737904532\"]\r\n<p id=\"fs-id1169737904532\">[latex]f^{\\prime}(x) = \\frac{5}{2(5x-7)}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737904560\" class=\"exercise\">\r\n<div id=\"fs-id1169737904562\" class=\"textbox\">\r\n<p id=\"fs-id1169737904564\"><strong>12.\u00a0<\/strong>[latex]f(x)=x^2 \\ln 9x[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737904626\" class=\"exercise\">\r\n<div id=\"fs-id1169737904628\" class=\"textbox\">\r\n<p id=\"fs-id1169737904630\"><strong>13.\u00a0<\/strong>[latex]f(x)=\\log(\\sec x)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169737904665\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737904665\"]\r\n<p id=\"fs-id1169737904665\">[latex]f^{\\prime}(x) = \\frac{\\tan x}{\\ln 10}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738054394\" class=\"exercise\">\r\n<div id=\"fs-id1169738054396\" class=\"textbox\">\r\n<p id=\"fs-id1169738054398\"><strong>14.\u00a0<\/strong>[latex]f(x)=\\log_7 (6x^4+3)^5[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738054489\" class=\"exercise\">\r\n<div id=\"fs-id1169738054491\" class=\"textbox\">\r\n<p id=\"fs-id1169738054493\"><strong>15.\u00a0<\/strong>[latex]f(x)=2^x \\cdot \\log_3 7^{x^2-4}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738054540\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738054540\"]\r\n<p id=\"fs-id1169738054540\">[latex]f^{\\prime}(x) = 2^x \\cdot \\ln 2 \\cdot \\log_3 7^{x^2-4} + 2^x \\cdot \\frac{2x \\ln 7}{\\ln 3}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169738199886\">For the following exercises, use logarithmic differentiation to find [latex]\\frac{dy}{dx}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169738199906\" class=\"exercise\">\r\n<div id=\"fs-id1169738199908\" class=\"textbox\">\r\n<p id=\"fs-id1169738199910\"><strong>16.\u00a0<\/strong>[latex]y=x^{\\sqrt{x}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738199977\" class=\"exercise\">\r\n<div id=\"fs-id1169738199979\" class=\"textbox\">\r\n<p id=\"fs-id1169738199981\"><strong>17.\u00a0<\/strong>[latex]y=(\\sin 2x)^{4x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738200016\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738200016\"]\r\n<p id=\"fs-id1169738200016\">[latex]\\frac{dy}{dx} = (\\sin 2x)^{4x} [4 \\cdot \\ln(\\sin 2x) + 8x \\cdot \\cot 2x][\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738092208\" class=\"exercise\">\r\n<div id=\"fs-id1169738092210\" class=\"textbox\">\r\n<p id=\"fs-id1169738092212\"><strong>18.\u00a0<\/strong>[latex]y=(\\ln x)^{\\ln x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738092313\" class=\"exercise\">\r\n<div id=\"fs-id1169738092315\" class=\"textbox\">\r\n<p id=\"fs-id1169738092318\"><strong>19.\u00a0<\/strong>[latex]y=x^{\\log_2 x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738092342\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738092342\"]\r\n<p id=\"fs-id1169738092342\">[latex]\\frac{dy}{dx} = x^{\\log_2 x} \\cdot \\frac{2 \\ln x}{x \\ln 2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738092392\" class=\"exercise\">\r\n<div id=\"fs-id1169738092394\" class=\"textbox\">\r\n<p id=\"fs-id1169738092396\"><strong>20.\u00a0<\/strong>[latex]y=(x^2-1)^{\\ln x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738039194\" class=\"exercise\">\r\n<div id=\"fs-id1169738039196\" class=\"textbox\">\r\n<p id=\"fs-id1169738039198\"><strong>21.\u00a0<\/strong>[latex]y=x^{\\cot x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738039222\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738039222\"]\r\n<p id=\"fs-id1169738039222\">[latex]\\frac{dy}{dx} = x^{\\cot x} \\cdot [\u2212\\csc^2 x \\cdot \\ln x+\\frac{\\cot x}{x}][\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738039285\" class=\"exercise\">\r\n<div id=\"fs-id1169738039287\" class=\"textbox\">\r\n<p id=\"fs-id1169738093910\"><strong>22.\u00a0<\/strong>[latex]y= \\large \\frac{x+11}{\\sqrt[3]{x^2-4}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738094022\" class=\"exercise\">\r\n<div id=\"fs-id1169738094024\" class=\"textbox\">\r\n<p id=\"fs-id1169738094026\"><strong>23.\u00a0<\/strong>[latex]y=x^{-1\/2}(x^2+3)^{2\/3}(3x-4)^4[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738094094\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738094094\"]\r\n<p id=\"fs-id1169738094094\">[latex]\\frac{dy}{dx} = x^{-1\/2}(x^2+3)^{2\/3}(3x-4)^4 \\cdot [\\frac{-1}{2x}+\\frac{4x}{3(x^2+3)}+\\frac{12}{3x-4}][\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738209953\" class=\"exercise\">\r\n<div id=\"fs-id1169738209955\" class=\"textbox\">\r\n<p id=\"fs-id1169738209958\"><strong>24. [T]<\/strong> Find an equation of the tangent line to the graph of [latex]f(x)=4xe^{x^2-1}[\/latex] at the point where<\/p>\r\n<p id=\"fs-id1169738210004\">[latex]x=-1[\/latex]. Graph both the function and the tangent line.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738210057\" class=\"exercise\">\r\n<div id=\"fs-id1169738210060\" class=\"textbox\">\r\n<p id=\"fs-id1169738210062\"><strong>25. [T]<\/strong> Find the equation of the line that is normal to the graph of [latex]f(x)=x \\cdot 5^x[\/latex] at the point where [latex]x=1[\/latex]. Graph both the function and the normal line.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738080236\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738080236\"]<span id=\"fs-id1169738080243\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205540\/CNX_Calc_Figure_03_09_202.jpg\" alt=\"The function starts at (\u22123, 0), decreases slightly and then increases through the origin and increases to (1.25, 10). There is a straight line marked T(x) with slope \u22121\/(5 + 5 ln 5) and y intercept 5 + 1\/(5 + 5 ln 5).\" \/><\/span>\r\n[latex]y=\\frac{-1}{5+5 \\ln 5}x+(5+\\frac{1}{5+5 \\ln 5})[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738080322\" class=\"exercise\">\r\n<div id=\"fs-id1169738080324\" class=\"textbox\">\r\n<p id=\"fs-id1169738080326\"><strong>26. [T]<\/strong> Find the equation of the tangent line to the graph of [latex]x^3-x \\ln y+y^3=2x+5[\/latex] at the point where [latex]x=2[\/latex]. (<em>Hint<\/em>: Use implicit differentiation to find [latex]\\frac{dy}{dx}[\/latex].) Graph both the curve and the tangent line.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738094644\" class=\"exercise\">\r\n<div id=\"fs-id1169738094646\" class=\"textbox\">\r\n<p id=\"fs-id1169738094648\"><strong>27.\u00a0<\/strong>Consider the function [latex]y=x^{1\/x}[\/latex] for [latex]x&gt;0[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1169738094680\" style=\"list-style-type: lower-alpha\">\r\n \t<li>Determine the points on the graph where the tangent line is horizontal.<\/li>\r\n \t<li>Determine the intervals where [latex]y^{\\prime}&gt;0[\/latex] and those where [latex]y^{\\prime}&lt;0[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738094723\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738094723\"]\r\n<p id=\"fs-id1169738094723\">a. [latex]x=e \\approx 2.718[\/latex]\r\nb. [latex]y^{\\prime}&gt;0[\/latex] on [latex](e,\\infty)[\/latex], and [latex]y^{\\prime}&lt;0[\/latex] on [latex](0,e)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738094772\" class=\"exercise\">\r\n<div id=\"fs-id1169738094774\" class=\"textbox\">\r\n<p id=\"fs-id1169738094776\"><strong>28.\u00a0<\/strong>The formula [latex]I(t)=\\frac{\\sin t}{e^t}[\/latex] is the formula for a decaying alternating current.<\/p>\r\n\r\n<ol id=\"fs-id1169738094810\" style=\"list-style-type: lower-alpha\">\r\n \t<li>Complete the following table with the appropriate values.\r\n<table id=\"fs-id1169738094824\" class=\"unnumbered\" summary=\"This table has two columns and 10 rows. The first column reads t, 0, \u03c0\/2, \u03c0, 3\u03c0\/2, 2\u03c0, 5\u03c0\/2, 3\u03c0, 7\u03c0\/2, and 4\u03c0. The second column reads (sin t)\/et, (i), (ii), (iii), (iv), (v), (vi), (vii), (viii), and (ix).\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>[latex]t[\/latex]<\/th>\r\n<th>[latex]\\frac{\\sin t}{e^t}[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>0<\/td>\r\n<td>(i)<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{\\pi}{2}[\/latex]<\/td>\r\n<td>(ii)<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\pi[\/latex]<\/td>\r\n<td>(iii)<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{3\\pi}{2}[\/latex]<\/td>\r\n<td>(iv)<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]2\\pi[\/latex]<\/td>\r\n<td>(v)<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]2\\pi[\/latex]<\/td>\r\n<td>(vi)<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]3\\pi[\/latex]<\/td>\r\n<td>(vii)<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]\\frac{7\\pi}{2}[\/latex]<\/td>\r\n<td>(viii)<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex]4\\pi[\/latex]<\/td>\r\n<td>(ix)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t<li>Using only the values in the table, determine where the tangent line to the graph of [latex]I(t)[\/latex] is horizontal.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738229480\" class=\"exercise\">\r\n<div id=\"fs-id1169738229482\" class=\"textbox\">\r\n<p id=\"fs-id1169738229484\"><strong>29. [T]<\/strong> The population of Toledo, Ohio, in 2000 was approximately 500,000. Assume the population is increasing at a rate of 5% per year.<\/p>\r\n\r\n<ol id=\"fs-id1169738229493\" style=\"list-style-type: lower-alpha\">\r\n \t<li>Write the exponential function that relates the total population as a function of [latex]t[\/latex].<\/li>\r\n \t<li>Use a. to determine the rate at which the population is increasing in [latex]t[\/latex] years.<\/li>\r\n \t<li>Use b. to determine the rate at which the population is increasing in 10 years.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738229522\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738229522\"]\r\n<p id=\"fs-id1169738229522\">a. [latex]P=500,000(1.05)^t[\/latex] individuals\r\nb. [latex]P^{\\prime}(t)=24395 \\cdot (1.05)^t[\/latex] individuals per year\r\nc. 39,737 individuals per year<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738229590\" class=\"exercise\">\r\n<div id=\"fs-id1169738229593\" class=\"textbox\">\r\n<p id=\"fs-id1169738229595\"><strong>30. [T]<\/strong> An isotope of the element erbium has a half-life of approximately 12 hours. Initially there are 9 grams of the isotope present.<\/p>\r\n\r\n<ol id=\"fs-id1169738229603\" style=\"list-style-type: lower-alpha\">\r\n \t<li>Write the exponential function that relates the amount of substance remaining as a function of [latex]t[\/latex], measured in hours.<\/li>\r\n \t<li>Use a. to determine the rate at which the substance is decaying in [latex]t[\/latex] hours.<\/li>\r\n \t<li>Use b. to determine the rate of decay at [latex]t=4[\/latex] hours.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738220789\" class=\"exercise\">\r\n<div id=\"fs-id1169738220792\" class=\"textbox\">\r\n<p id=\"fs-id1169738220794\"><strong>31. [T]<\/strong> The number of cases of influenza in New York City from the beginning of 1960 to the beginning of 1961 is modeled by the function<\/p>\r\n<p id=\"fs-id1169738220802\">[latex]N(t)=5.3e^{0.093t^2-0.87t}, \\, (0\\le t\\le 4)[\/latex],<\/p>\r\n<p id=\"fs-id1169738220862\">where [latex]N(t)[\/latex] gives the number of cases (in thousands) and [latex]t[\/latex] is measured in years, with [latex]t=0[\/latex] corresponding to the beginning of 1960.<\/p>\r\n\r\n<ol id=\"fs-id1169738220892\" style=\"list-style-type: lower-alpha\">\r\n \t<li>Show work that evaluates [latex]N(0)[\/latex] and [latex]N(4)[\/latex]. Briefly describe what these values indicate about the disease in New York City.<\/li>\r\n \t<li>Show work that evaluates [latex]N^{\\prime}(0)[\/latex] and [latex]N^{\\prime}(3)[\/latex]. Briefly describe what these values indicate about the disease in New York City.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169737922881\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169737922881\"]\r\n<p id=\"fs-id1169737922881\">a. At the beginning of 1960 there were 5.3 thousand cases of the disease in New York City. At the beginning of 1963 there were approximately 723 cases of the disease in New York City.\r\nb. At the beginning of 1960 the number of cases of the disease was decreasing at rate of -4.611 thousand per year; at the beginning of 1963, the number of cases of the disease was decreasing at a rate of -0.2808 thousand per year.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737922901\" class=\"exercise\">\r\n<div id=\"fs-id1169737922903\" class=\"textbox\">\r\n<p id=\"fs-id1169737922905\"><strong>32. [T]<\/strong> The <em>relative rate of change<\/em> of a differentiable function [latex]y=f(x)[\/latex] is given by [latex]\\frac{100 \\cdot f^{\\prime}(x)}{f(x)}\\%[\/latex]. One model for population growth is a Gompertz growth function, given by [latex]P(x)=ae^{\u2212b \\cdot e^{\u2212cx}}[\/latex] where [latex]a, \\, b[\/latex], and [latex]c[\/latex] are constants.<\/p>\r\n\r\n<ol id=\"fs-id1169737923026\" style=\"list-style-type: lower-alpha\">\r\n \t<li>Find the relative rate of change formula for the generic Gompertz function.<\/li>\r\n \t<li>Use a. to find the relative rate of change of a population in [latex]x=20[\/latex] months when [latex]a=204,b=0.0198,[\/latex] and [latex]c=0.15.[\/latex]<\/li>\r\n \t<li>Briefly interpret what the result of b. means.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169738227093\">For the following exercises, use the population of New York City from 1790 to 1860, given in the following table.<\/p>\r\n\r\n<table id=\"fs-id1169738227097\" summary=\"This table has nine rows and two columns. The first row is a header row and it labels each column. The first column header is Years since 1790 and the second column is Population. Under the first column are the values 0, 10, 20, 30, 40, 50, 60, and 70. Under the second column are the values 33,131; 60,515; 96,373; 123,706; 202,300; 312,710; 515,547; and 813,669.\"><caption>New York City Population Over Time\r\nSource: http:\/\/en.wikipedia.org\/wiki\/Largest_cities_in_the_United_States_by_population_by_decade<\/caption><\/table>\r\n<table style=\"border-collapse: collapse;width: 100%\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 50%;text-align: center\"><strong>Years since 1790<\/strong><\/td>\r\n<td style=\"width: 50%;text-align: center\"><strong>Population<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;text-align: center\">0<\/td>\r\n<td style=\"width: 50%;text-align: center\">33,131<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;text-align: center\">10<\/td>\r\n<td style=\"width: 50%;text-align: center\">60,515<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;text-align: center\">20<\/td>\r\n<td style=\"width: 50%;text-align: center\">96,373<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;text-align: center\">30<\/td>\r\n<td style=\"width: 50%;text-align: center\">123,706<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;text-align: center\">40<\/td>\r\n<td style=\"width: 50%;text-align: center\">202,300<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;text-align: center\">50<\/td>\r\n<td style=\"width: 50%;text-align: center\">312,710<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;text-align: center\">60<\/td>\r\n<td style=\"width: 50%;text-align: center\">515,547<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;text-align: center\">70<\/td>\r\n<td style=\"width: 50%;text-align: center\">813,669<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-id1169738066609\" class=\"exercise\">\r\n<div id=\"fs-id1169738066611\" class=\"textbox\">\r\n<p id=\"fs-id1169738066613\"><strong>33. [T]<\/strong> Using a computer program or a calculator, fit a growth curve to the data of the form [latex]p=ab^t[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738066639\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738066639\"]\r\n<p id=\"fs-id1169738066639\">[latex]p=35741(1.045)^t[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738066665\" class=\"exercise\">\r\n<div id=\"fs-id1169738066667\" class=\"textbox\">\r\n<p id=\"fs-id1169738066669\"><strong>34. [T]<\/strong> Using the exponential best fit for the data, write a table containing the derivatives evaluated at each year.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738071156\" class=\"exercise\">\r\n<div id=\"fs-id1169738071158\" class=\"textbox\">\r\n\r\n<strong>35. [T]<\/strong> Using the exponential best fit for the data, write a table containing the second derivatives evaluated at each year.\r\n\r\n<\/div>\r\n<div id=\"fs-id1169738071170\" class=\"textbox shaded\">[reveal-answer q=\"fs-id1169738071170\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738071170\"]\r\n<table id=\"fs-id1169738071176\" class=\"unnumbered\" summary=\"This table has nine rows and two columns. The first row is a header row and it labels each column. The first column header is Years since 1790 and the second column is P\u2019\u2019. Under the first column are the values 0, 10, 20, 30, 40, 50, 60, and 70. Under the second column are the values 69.25, 107.5, 167.0, 259.4, 402.8, 625.5, 971.4, and 1508.5.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th style=\"text-align: center\">Years since 1790<\/th>\r\n<th style=\"text-align: center\">[latex]P''[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td style=\"text-align: center\">0<\/td>\r\n<td style=\"text-align: center\">69.25<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"text-align: center\">10<\/td>\r\n<td style=\"text-align: center\">107.5<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"text-align: center\">20<\/td>\r\n<td style=\"text-align: center\">167.0<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"text-align: center\">30<\/td>\r\n<td style=\"text-align: center\">259.4<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"text-align: center\">40<\/td>\r\n<td style=\"text-align: center\">402.8<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"text-align: center\">50<\/td>\r\n<td style=\"text-align: center\">625.5<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"text-align: center\">60<\/td>\r\n<td style=\"text-align: center\">971.4<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"text-align: center\">70<\/td>\r\n<td style=\"text-align: center\">1508.5<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738040664\" class=\"exercise\">\r\n<div id=\"fs-id1169738040666\" class=\"textbox\">\r\n<p id=\"fs-id1169738040668\"><strong>36. [T]<\/strong> Using the tables of first and second derivatives and the best fit, answer the following questions:<\/p>\r\n\r\n<ol id=\"fs-id1169738040677\" style=\"list-style-type: lower-alpha\">\r\n \t<li>Will the model be accurate in predicting the future population of New York City? Why or why not?<\/li>\r\n \t<li>Estimate the population in 2010. Was the prediction correct from a.?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1169738074914\" class=\"definition\">\r\n \t<dt>logarithmic differentiation<\/dt>\r\n \t<dd id=\"fs-id1169738074919\">is a technique that allows us to differentiate a function by first taking the natural logarithm of both sides of an equation, applying properties of logarithms to simplify the equation, and differentiating implicitly<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Find the derivative of exponential functions.<\/li>\n<li>Find the derivative of logarithmic functions.<\/li>\n<li>Use logarithmic differentiation to determine the derivative of a function.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1169738221270\">So far, we have learned how to differentiate a variety of functions, including trigonometric, inverse, and implicit functions. In this section, we explore derivatives of exponential and logarithmic functions. As we discussed in <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/introduction\/\">Introduction to Functions and Graphs<\/a>, exponential functions play an important role in modeling population growth and the decay of radioactive materials. Logarithmic functions can help rescale large quantities and are particularly helpful for rewriting complicated expressions.<\/p>\n<div id=\"fs-id1169737795671\" class=\"bc-section section\">\n<h1>Derivative of the Exponential Function<\/h1>\n<p id=\"fs-id1169738221359\">Just as when we found the derivatives of other functions, we can find the derivatives of exponential and logarithmic functions using formulas. As we develop these formulas, we need to make certain basic assumptions. The proofs that these assumptions hold are beyond the scope of this course.<\/p>\n<p id=\"fs-id1169738045927\">First of all, we begin with the assumption that the function [latex]B(x)=b^x, \\, b>0[\/latex], is defined for every real number and is continuous. In previous courses, the values of exponential functions for all rational numbers were defined\u2014beginning with the definition of [latex]b^n[\/latex], where [latex]n[\/latex] is a positive integer\u2014as the product of [latex]b[\/latex] multiplied by itself [latex]n[\/latex] times. Later, we defined [latex]b^0=1, \\, b^{\u2212n}=\\frac{1}{b^n}[\/latex] for a positive integer [latex]n[\/latex], and [latex]b^{s\/t}=(\\sqrt[t]{b})^s[\/latex] for positive integers [latex]s[\/latex] and [latex]t[\/latex]. These definitions leave open the question of the value of [latex]b^r[\/latex] where [latex]r[\/latex] is an arbitrary real number. By assuming the <em>continuity<\/em> of [latex]B(x)=b^x, \\, b>0[\/latex], we may interpret [latex]b^r[\/latex] as [latex]\\underset{x\\to r}{\\lim}b^x[\/latex] where the values of [latex]x[\/latex] as we take the limit are rational. For example, we may view [latex]{4}^{\\pi}[\/latex] as the number satisfying<\/p>\n<div id=\"fs-id1169738212703\" class=\"equation unnumbered\">[latex]\\begin{array}{l}4^3<4^{\\pi}<4^4, \\, 4^{3.1}<4^{\\pi}<4^{3.2}, \\, 4^{3.14}<4^{\\pi}<4^{3.15},\\\\ 4^{3.141}<4^{\\pi}<4^{3.142}, \\, 4^{3.1415}<4^{\\pi}<4^{3.1416}, \\, \\cdots \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1169737765653\">As we see in the following table, [latex]4^{\\pi}\\approx 77.88[\/latex].<\/p>\n<table id=\"fs-id1169738222068\" summary=\"This table has seven rows and four columns. The first row is a header row and it labels each column. The first column header is x, the second column header is 4x, the third column header is x, and the fourth column header is 4x. Under the first column are the values 43, 43.1, 43.14, 43.141, 43.1415. Under the second column are the values 64, 73.5166947198, 77.7084726013, 77.8162741237, 77.8702309526, 77.8799471543. Under the third column are the values 43.141593, 43.1416, 43.142, 43.15, 43.2, and 44. Under the fourth column are the values 77.8802710486, 77.8810268071, 77.9242251944, 78.7932424541, 84.4485062895, and 256.\">\n<caption>Approximating a Value of [latex]4^{\\pi}[\/latex]<\/caption>\n<thead>\n<tr valign=\"top\">\n<th>[latex]x[\/latex]<\/th>\n<th>[latex]4^x[\/latex]<\/th>\n<th>[latex]x[\/latex]<\/th>\n<th>[latex]4^x[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex]4^3[\/latex]<\/td>\n<td>64<\/td>\n<td>[latex]4^{3.141593}[\/latex]<\/td>\n<td>77.8802710486<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]4^{3.1}[\/latex]<\/td>\n<td>73.5166947198<\/td>\n<td>[latex]4^{3.1416}[\/latex]<\/td>\n<td>77.8810268071<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]4^{3.14}[\/latex]<\/td>\n<td>77.7084726013<\/td>\n<td>[latex]4^{3.142}[\/latex]<\/td>\n<td>77.9242251944<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]4^{3.141}[\/latex]<\/td>\n<td>77.8162741237<\/td>\n<td>[latex]4^{3.15}[\/latex]<\/td>\n<td>78.7932424541<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]4^{3.1415}[\/latex]<\/td>\n<td>77.8702309526<\/td>\n<td>[latex]4^{3.2}[\/latex]<\/td>\n<td>84.4485062895<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]4^{3.14159}[\/latex]<\/td>\n<td>77.8799471543<\/td>\n<td>[latex]4^4[\/latex]<\/td>\n<td>256<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1169737707453\">We also assume that for [latex]B(x)=b^x, \\, b>0[\/latex], the value [latex]B^{\\prime}(0)[\/latex] of the derivative exists. In this section, we show that by making this one additional assumption, it is possible to prove that the function [latex]B(x)[\/latex] is differentiable everywhere.<\/p>\n<p id=\"fs-id1169737728646\">We make one final assumption: that there is a unique value of [latex]b>0[\/latex] for which [latex]B^{\\prime}(0)=1[\/latex]. We define [latex]e[\/latex] to be this unique value, as we did in <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/chapter\/introduction\/\">Introduction to Functions and Graphs<\/a>. <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_03_09_001\">(Figure)<\/a> provides graphs of the functions [latex]y=2^x, \\, y=3^x, \\, y=2.7^x[\/latex], and [latex]y=2.8^x[\/latex]. A visual estimate of the slopes of the tangent lines to these functions at 0 provides evidence that the value of [latex]e[\/latex] lies somewhere between 2.7 and 2.8. The function [latex]E(x)=e^x[\/latex] is called the <strong>natural exponential function<\/strong>. Its inverse, [latex]L(x)=\\log_e x=\\ln x[\/latex] is called the <strong>natural logarithmic function<\/strong>.<\/p>\n<div id=\"CNX_Calc_Figure_03_09_001\" class=\"wp-caption aligncenter\">\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205529\/CNX_Calc_Figure_03_09_001.jpg\" alt=\"The graphs of 3x, 2.8x, 2.7x, and 2x are shown. In quadrant I, their order from least to greatest is 2x, 2.7x, 2.8x, and 3x. In quadrant II, this order is reversed. All cross the y-axis at (0, 1).\" width=\"731\" height=\"592\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 1.<\/strong> The graph of [latex]E(x)=e^x[\/latex] is between [latex]y=2^x[\/latex] and [latex]y=3^x[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<p id=\"fs-id1169737717318\">For a better estimate of [latex]e[\/latex], we may construct a table of estimates of [latex]B^{\\prime}(0)[\/latex] for functions of the form [latex]B(x)=b^x[\/latex]. Before doing this, recall that<\/p>\n<div id=\"fs-id1169738116003\" class=\"equation unnumbered\">[latex]B^{\\prime}(0)=\\underset{x\\to 0}{\\lim}\\frac{b^x-b^0}{x-0}=\\underset{x\\to 0}{\\lim}\\frac{b^x-1}{x} \\approx \\frac{b^x-1}{x}[\/latex]<\/div>\n<p id=\"fs-id1169737790576\">for values of [latex]x[\/latex] very close to zero. For our estimates, we choose [latex]x=0.00001[\/latex] and [latex]x=-0.00001[\/latex] to obtain the estimate<\/p>\n<div id=\"fs-id1169737725568\" class=\"equation unnumbered\">[latex]\\frac{b^{-0.00001}-1}{-0.00001}<B^{\\prime}(0)<\\frac{b^{0.00001}-1}{0.00001}[\/latex].<\/div>\n<p id=\"fs-id1169737952624\">See the following table.<\/p>\n<table id=\"fs-id1169738019199\" summary=\"This table has six rows and four columns. The first row is a header row and it labels each column. The first column header is b, the second column header is (b\u22120.00001 \u2212 1)\/\u22120.00001 &lt; B\u2019(0) &lt; (b0.00001 \u2212 1)\/0.00001, the third column header is b, and the fourth column header is (b\u22120.00001 \u2212 1)\/\u22120.00001 &lt; B\u2019(0) &lt; (b0.00001 \u2212 1)\/0.00001. Under the first column are the values 2, 2.7, 2.71, 2.718, and 2.7182. Under the second column are the values 0.693145&lt;B\u2019(0)&lt;0.69315, 0.993247&lt;B\u2019(0)&lt; 0.993257, 0.996944&lt;B\u2019(0)&lt;0.996954, 0.999891&lt;B\u2019(0)&lt; 0.999901, and 0.999965&lt;B\u2019(0)&lt;0.999975. Under the third column are the values 2.7183, 2.719, 2.72, 2.8, and 3. Under the fourth column are the values 1.000002&lt;B\u2019(0)&lt; 1.000012, 1.000259&lt;B\u2019(0)&lt; 1.000269, 1.000627&lt;B\u2019(0)&lt;1.000637, 1.029614&lt;B\u2019(0)&lt;1.029625, and 1.098606&lt;B\u2019(00&lt;1.098618.\">\n<caption>Estimating a Value of [latex]e[\/latex]<\/caption>\n<thead>\n<tr valign=\"top\">\n<th>[latex]b[\/latex]<\/th>\n<th>[latex]\\frac{b^{-0.00001}-1}{-0.00001}<B^{\\prime}(0)<\\frac{b^{0.00001}-1}{0.00001}[\/latex]<\/th>\n<th>[latex]b[\/latex]<\/th>\n<th>[latex]\\frac{b^{-0.00001}-1}{-0.00001}<B^{\\prime}(0)<\\frac{b^{0.00001}-1}{0.00001}[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>2<\/td>\n<td>[latex]0.693145<B^{\\prime}(0)<0.69315[\/latex]<\/td>\n<td>2.7183<\/td>\n<td>[latex]1.000002<B^{\\prime}(0)<1.000012[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2.7<\/td>\n<td>[latex]0.993247<B^{\\prime}(0)<0.993257[\/latex]<\/td>\n<td>2.719<\/td>\n<td>[latex]1.000259<B^{\\prime}(0)<1.000269[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2.71<\/td>\n<td>[latex]0.996944<B^{\\prime}(0)<0.996954[\/latex]<\/td>\n<td>2.72<\/td>\n<td>[latex]1.000627<B^{\\prime}(0)<1.000637[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2.718<\/td>\n<td>[latex]0.999891<B^{\\prime}(0)<0.999901[\/latex]<\/td>\n<td>2.8<\/td>\n<td>[latex]1.029614<B^{\\prime}(0)<1.029625[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>2.7182<\/td>\n<td>[latex]0.999965<B^{\\prime}(0)<0.999975[\/latex]<\/td>\n<td>3<\/td>\n<td>[latex]1.098606<B^{\\prime}(0)<1.098618[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1169738213815\">The evidence from the table suggests that [latex]2.7182<e<2.7183[\/latex].<\/p>\n<p id=\"fs-id1169737978597\">The graph of [latex]E(x)=e^x[\/latex] together with the line [latex]y=x+1[\/latex] are shown in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_03_09_002\">(Figure)<\/a>. This line is tangent to the graph of [latex]E(x)=e^x[\/latex] at [latex]x=0[\/latex].<\/p>\n<div id=\"CNX_Calc_Figure_03_09_002\" class=\"wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205533\/CNX_Calc_Figure_03_09_002.jpg\" alt=\"Graph of the function ex along with its tangent at (0, 1), x + 1.\" width=\"487\" height=\"248\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 2.<\/strong> The tangent line to [latex]E(x)=e^x[\/latex] at [latex]x=0[\/latex] has slope 1.<\/p>\n<\/div>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<p id=\"fs-id1169738198749\">Now that we have laid out our basic assumptions, we begin our investigation by exploring the derivative of [latex]B(x)=b^x, \\, b>0[\/latex]. Recall that we have assumed that [latex]B^{\\prime}(0)[\/latex] exists. By applying the limit definition to the derivative we conclude that<\/p>\n<div id=\"fs-id1169738045860\" class=\"equation\">[latex]B^{\\prime}(0)=\\underset{h\\to 0}{\\lim}\\frac{b^{0+h}-b^0}{h}=\\underset{h\\to 0}{\\lim}\\frac{b^h-1}{h}[\/latex].<\/div>\n<p id=\"fs-id1169738187820\">Turning to [latex]B^{\\prime}(x)[\/latex], we obtain the following.<\/p>\n<div id=\"fs-id1169737954073\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} B^{\\prime}(x) & =\\underset{h\\to 0}{\\lim}\\frac{b^{x+h}-b^x}{h} & & & \\text{Apply the limit definition of the derivative.} \\\\ & =\\underset{h\\to 0}{\\lim}\\frac{b^xb^h-b^x}{h} & & & \\text{Note that} \\, b^{x+h}=b^x b^h. \\\\ & =\\underset{h\\to 0}{\\lim}\\frac{b^x(b^h-1)}{h} & & & \\text{Factor out} \\, b^x. \\\\ & =b^x\\underset{h\\to 0}{\\lim}\\frac{b^h-1}{h} & & & \\text{Apply a property of limits.} \\\\ & =b^x B^{\\prime}(0) & & & \\text{Use} \\, B^{\\prime}(0)=\\underset{h\\to 0}{\\lim}\\frac{b^{0+h}-b^0}{h}=\\underset{h\\to 0}{\\lim}\\frac{b^h-1}{h}. \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1169738221160\">We see that on the basis of the assumption that [latex]B(x)=b^x[\/latex] is differentiable at [latex]0, \\, B(x)[\/latex] is not only differentiable everywhere, but its derivative is<\/p>\n<div id=\"fs-id1169738221385\" class=\"equation\">[latex]B^{\\prime}(x)=b^x B^{\\prime}(0)[\/latex].<\/div>\n<p id=\"fs-id1169738183261\">For [latex]E(x)=e^x, \\, E^{\\prime}(0)=1[\/latex]. Thus, we have [latex]E^{\\prime}(x)=e^x[\/latex]. (The value of [latex]B^{\\prime}(0)[\/latex] for an arbitrary function of the form [latex]B(x)=b^x, \\, b>0[\/latex], will be derived later.)<\/p>\n<div id=\"fs-id1169738226753\" class=\"textbox key-takeaways theorem\">\n<h3>Derivative of the Natural Exponential Function<\/h3>\n<p id=\"fs-id1169738220224\">Let [latex]E(x)=e^x[\/latex] be the natural exponential function. Then<\/p>\n<div id=\"fs-id1169737928243\" class=\"equation unnumbered\">[latex]E^{\\prime}(x)=e^x[\/latex].<\/div>\n<p id=\"fs-id1169737142141\">In general,<\/p>\n<div id=\"fs-id1169738124964\" class=\"equation unnumbered\">[latex]\\frac{d}{dx}(e^{g(x)})=e^{g(x)} g^{\\prime}(x)[\/latex].<\/div>\n<\/div>\n<div id=\"fs-id1169738223456\" class=\"textbox examples\">\n<h3>Derivative of an Exponential Function<\/h3>\n<div id=\"fs-id1169738223458\" class=\"exercise\">\n<div id=\"fs-id1169738223460\" class=\"textbox\">\n<p id=\"fs-id1169738187154\">Find the derivative of [latex]f(x)=e^{\\tan (2x)}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737140844\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737140844\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737140844\">Using the derivative formula and the chain rule,<\/p>\n<div id=\"fs-id1169738048872\" class=\"equation unnumbered\">[latex]\\begin{array}{ll} f^{\\prime}(x) & =e^{\\tan (2x)}\\frac{d}{dx}(\\tan (2x)) \\\\ & = e^{\\tan (2x)} \\sec^2 (2x) \\cdot 2. \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737140879\" class=\"textbox examples\">\n<h3>Combining Differentiation Rules<\/h3>\n<div id=\"fs-id1169737140881\" class=\"exercise\">\n<div id=\"fs-id1169737766541\" class=\"textbox\">\n<p id=\"fs-id1169737766547\">Find the derivative of [latex]y=\\frac{e^{x^2}}{x}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737928258\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737928258\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737928258\">Use the derivative of the natural exponential function, the quotient rule, and the chain rule.<\/p>\n<div id=\"fs-id1169737928262\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} y^{\\prime} & =\\large \\frac{(e^{x^2} \\cdot 2x) \\cdot x - 1 \\cdot e^{x^2}}{x^2} & & & \\text{Apply the quotient rule.} \\\\ & = \\large \\frac{e^{x^2}(2x^2-1)}{x^2} & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738152550\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1169738152553\" class=\"exercise\">\n<div id=\"fs-id1169737923849\" class=\"textbox\">\n<p id=\"fs-id1169737923851\">Find the derivative of [latex]h(x)=xe^{2x}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737948362\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737948362\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737948362\">[latex]h^{\\prime}(x)=e^{2x}+2xe^{2x}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1169738222020\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1169738215078\">Don\u2019t forget to use the product rule.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737949367\" class=\"textbox examples\">\n<h3>Applying the Natural Exponential Function<\/h3>\n<div id=\"fs-id1169737949369\" class=\"exercise\">\n<div id=\"fs-id1169737151938\" class=\"textbox\">\n<p id=\"fs-id1169737151943\">A colony of mosquitoes has an initial population of 1000. After [latex]t[\/latex] days, the population is given by [latex]A(t)=1000e^{0.3t}[\/latex]. Show that the ratio of the rate of change of the population, [latex]A^{\\prime}(t)[\/latex], to the population size, [latex]A(t)[\/latex] is constant.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738212487\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738212487\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738212487\">First find [latex]A^{\\prime}(t)[\/latex]. By using the chain rule, we have [latex]A^{\\prime}(t)=300e^{0.3t}[\/latex]. Thus, the ratio of the rate of change of the population to the population size is given by<\/p>\n<div id=\"fs-id1169738227689\" class=\"equation unnumbered\">[latex]\\large \\frac{A^{\\prime}(t)}{A(t)} \\normalsize = \\large \\frac{300e^{0.3t}}{1000e^{0.3t}}=0.3[\/latex].<\/div>\n<p id=\"fs-id1169737145031\">The ratio of the rate of change of the population to the population size is the constant 0.3.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738186819\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1169738186822\" class=\"exercise\">\n<div id=\"fs-id1169738186824\" class=\"textbox\">\n<p id=\"fs-id1169738215955\">If [latex]A(t)=1000e^{0.3t}[\/latex] describes the mosquito population after [latex]t[\/latex] days, as in the preceding example, what is the rate of change of [latex]A(t)[\/latex] after 4 days?<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737934408\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737934408\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737934408\">996<\/p>\n<\/div>\n<div class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1169738225603\">Find [latex]A^{\\prime}(4)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738221999\" class=\"bc-section section\">\n<h1>Derivative of the Logarithmic Function<\/h1>\n<p id=\"fs-id1169738222225\">Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function.<\/p>\n<div id=\"fs-id1169737927590\" class=\"textbox key-takeaways theorem\">\n<h3>The Derivative of the Natural Logarithmic Function<\/h3>\n<p id=\"fs-id1169737911417\">If [latex]x>0[\/latex] and [latex]y=\\ln x[\/latex], then<\/p>\n<div id=\"fs-id1169738223534\" class=\"equation\">[latex]\\frac{dy}{dx}=\\frac{1}{x}[\/latex].<\/div>\n<p id=\"fs-id1169737765282\">More generally, let [latex]g(x)[\/latex] be a differentiable function. For all values of [latex]x[\/latex] for which [latex]g^{\\prime}(x)>0[\/latex], the derivative of [latex]h(x)=\\ln(g(x))[\/latex] is given by<\/p>\n<div id=\"fs-id1169737919348\" class=\"equation\">[latex]h^{\\prime}(x)=\\frac{1}{g(x)} g^{\\prime}(x)[\/latex].<\/div>\n<\/div>\n<div id=\"fs-id1169738093053\" class=\"bc-section section\">\n<h2>Proof<\/h2>\n<p id=\"fs-id1169738186955\">If [latex]x>0[\/latex] and [latex]y=\\ln x[\/latex], then [latex]e^y=x[\/latex]. Differentiating both sides of this equation results in the equation<\/p>\n<div id=\"fs-id1169738212688\" class=\"equation unnumbered\">[latex]e^y\\frac{dy}{dx}=1[\/latex].<\/div>\n<p id=\"fs-id1169738216993\">Solving for [latex]\\frac{dy}{dx}[\/latex] yields<\/p>\n<div class=\"equation unnumbered\">[latex]\\frac{dy}{dx}=\\frac{1}{e^y}[\/latex].<\/div>\n<p id=\"fs-id1169738219487\">Finally, we substitute [latex]x=e^y[\/latex] to obtain<\/p>\n<div id=\"fs-id1169737145231\" class=\"equation unnumbered\">[latex]\\frac{dy}{dx}=\\frac{1}{x}[\/latex].<\/div>\n<p id=\"fs-id1169738221909\">We may also derive this result by applying the inverse function theorem, as follows. Since [latex]y=g(x)=\\ln x[\/latex] is the inverse of [latex]f(x)=e^x[\/latex], by applying the inverse function theorem we have<\/p>\n<div id=\"fs-id1169738070944\" class=\"equation unnumbered\">[latex]\\frac{dy}{dx}=\\frac{1}{f^{\\prime}(g(x))}=\\frac{1}{e^{\\ln x}}=\\frac{1}{x}[\/latex].<\/div>\n<p id=\"fs-id1169737951681\">Using this result and applying the chain rule to [latex]h(x)=\\ln(g(x))[\/latex] yields<\/p>\n<div id=\"fs-id1169738106180\" class=\"equation unnumbered\">[latex]h^{\\prime}(x)=\\frac{1}{g(x)} g^{\\prime}(x). _\\blacksquare[\/latex]<\/div>\n<p>The graph of [latex]y=\\ln x[\/latex] and its derivative [latex]\\frac{dy}{dx}=\\frac{1}{x}[\/latex] are shown in <a class=\"autogenerated-content\" href=\"#CNX_Calc_Figure_03_09_003\">(Figure)<\/a>.<\/p>\n<div id=\"CNX_Calc_Figure_03_09_003\" class=\"wp-caption aligncenter\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205536\/CNX_Calc_Figure_03_09_003.jpg\" alt=\"Graph of the function ln x along with its derivative 1\/x. The function ln x is increasing on (0, + \u221e). Its derivative is decreasing but greater than 0 on (0, + \u221e).\" width=\"487\" height=\"209\" \/><\/p>\n<p class=\"wp-caption-text\"><strong> Figure 3.<\/strong>\u00a0The function [latex]y=\\ln x[\/latex] is increasing on [latex](0,+\\infty)[\/latex]. Its derivative [latex]y^{\\prime} =\\frac{1}{x}[\/latex] is greater than zero on [latex](0,+\\infty)[\/latex].<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738211098\" class=\"textbox examples\">\n<h3>Taking a Derivative of a Natural Logarithm<\/h3>\n<div id=\"fs-id1169738228370\" class=\"exercise\">\n<div id=\"fs-id1169738228372\" class=\"textbox\">\n<p id=\"fs-id1169738228377\">Find the derivative of [latex]f(x)=\\ln(x^3+3x-4)[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738221312\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738221312\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738221312\">Use <a class=\"autogenerated-content\" href=\"#fs-id1169737919348\">(Figure)<\/a> directly.<\/p>\n<div id=\"fs-id1169738220266\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} f^{\\prime}(x) & =\\frac{1}{x^3+3x-4} \\cdot (3x^2+3) & & & \\text{Use} \\, g(x)=x^3+3x-4 \\, \\text{in} \\, h^{\\prime}(x)=\\frac{1}{g(x)} g^{\\prime}(x). \\\\ & =\\frac{3x^2+3}{x^3+3x-4} & & & \\text{Rewrite.} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738244488\" class=\"textbox examples\">\n<h3>Using Properties of Logarithms in a Derivative<\/h3>\n<div id=\"fs-id1169738106136\" class=\"exercise\">\n<div id=\"fs-id1169738106138\" class=\"textbox\">\n<p id=\"fs-id1169738106144\">Find the derivative of [latex]f(x)=\\ln(\\frac{x^2 \\sin x}{2x+1})[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738219674\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738219674\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738219674\">At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.<\/p>\n<div id=\"fs-id1169738219679\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} f(x) & = \\ln(\\frac{x^2 \\sin x}{2x+1})=2\\ln x+\\ln(\\sin x)-\\ln(2x+1) & & & \\text{Apply properties of logarithms.} \\\\ f^{\\prime}(x) & = \\frac{2}{x} + \\frac{\\cos x}{\\sin x} -\\frac{2}{2x+1} & & & \\text{Apply sum rule and} \\, h^{\\prime}(x)=\\frac{1}{g(x)} g^{\\prime}(x). \\\\ & = \\frac{2}{x} + \\cot x - \\frac{2}{2x+1} & & & \\text{Simplify using the quotient identity for cotangent.} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738219654\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1169738219657\" class=\"exercise\">\n<div id=\"fs-id1169738219659\" class=\"textbox\">\n<p id=\"fs-id1169738219661\">Differentiate: [latex]f(x)=\\ln (3x+2)^5[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738192196\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738192196\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738192196\">[latex]f^{\\prime}(x)=\\frac{15}{3x+2}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1169738073195\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1169738073202\">Use a property of logarithms to simplify before taking the derivative.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169738076342\">Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of [latex]y=\\log_b x[\/latex] and [latex]y=b^x[\/latex] for [latex]b>0, \\, b\\ne 1[\/latex].<\/p>\n<div id=\"fs-id1169738238181\" class=\"textbox key-takeaways theorem\">\n<h3>Derivatives of General Exponential and Logarithmic Functions<\/h3>\n<p id=\"fs-id1169738186170\">Let [latex]b>0, \\, b\\ne 1[\/latex], and let [latex]g(x)[\/latex] be a differentiable function.<\/p>\n<ol id=\"fs-id1169737998025\">\n<li>If [latex]y=\\log_b x[\/latex], then\n<div id=\"fs-id1169738105090\" class=\"equation\">[latex]\\frac{dy}{dx}=\\frac{1}{x \\ln b}[\/latex].<\/div>\n<p>More generally, if [latex]h(x)=\\log_b (g(x))[\/latex], then for all values of [latex]x[\/latex] for which [latex]g(x)>0[\/latex],<\/p>\n<div id=\"fs-id1169738186308\" class=\"equation\">[latex]h^{\\prime}(x)=\\frac{g^{\\prime}(x)}{g(x) \\ln b}[\/latex].<\/p>\n<\/div>\n<\/li>\n<li>If [latex]y=b^x[\/latex], then\n<div id=\"fs-id1169738224034\" class=\"equation\">[latex]\\frac{dy}{dx}=b^x \\ln b[\/latex].<\/div>\n<p>More generally, if [latex]h(x)=b^{g(x)}[\/latex], then<\/p>\n<div id=\"fs-id1169738045159\" class=\"equation\">[latex]h^{\\prime}(x)=b^{g(x)} g^{\\prime}(x) \\ln b[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div class=\"bc-section section\">\n<h2>Proof<\/h2>\n<p id=\"fs-id1169738184804\">If [latex]y=\\log_b x[\/latex], then [latex]b^y=x[\/latex]. It follows that [latex]\\ln(b^y)=\\ln x[\/latex]. Thus [latex]y \\ln b = \\ln x[\/latex]. Solving for [latex]y[\/latex], we have [latex]y=\\frac{\\ln x}{\\ln b}[\/latex]. Differentiating and keeping in mind that [latex]\\ln b[\/latex] is a constant, we see that<\/p>\n<div id=\"fs-id1169738211798\" class=\"equation unnumbered\">[latex]\\frac{dy}{dx}=\\frac{1}{x \\ln b}[\/latex].<\/div>\n<p id=\"fs-id1169738211864\">The derivative in <a class=\"autogenerated-content\" href=\"#fs-id1169738186308\">(Figure)<\/a> now follows from the chain rule.<\/p>\n<p id=\"fs-id1169737954089\">If [latex]y=b^x[\/latex], then [latex]\\ln y=x \\ln b[\/latex]. Using implicit differentiation, again keeping in mind that [latex]\\ln b[\/latex] is constant, it follows that [latex]\\frac{1}{y}\\frac{dy}{dx}=\\text{ln}b.[\/latex] Solving for [latex]\\frac{dy}{dx}[\/latex] and substituting [latex]y=b^x[\/latex], we see that<\/p>\n<div id=\"fs-id1169737934328\" class=\"equation unnumbered\">[latex]\\frac{dy}{dx}=y \\ln b=b^x \\ln b[\/latex].<\/div>\n<p id=\"fs-id1169737145057\">The more general derivative (<a class=\"autogenerated-content\" href=\"#fs-id1169738045159\">(Figure)<\/a>) follows from the chain rule. [latex]_\\blacksquare[\/latex]<\/p>\n<div id=\"fs-id1169737145066\" class=\"textbox examples\">\n<h3>Applying Derivative Formulas<\/h3>\n<div class=\"exercise\">\n<div id=\"fs-id1169738185254\" class=\"textbox\">\n<p id=\"fs-id1169738185259\">Find the derivative of [latex]h(x)=\\large \\frac{3^x}{3^x+2}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737700313\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737700313\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737700313\">Use the quotient rule and <a class=\"autogenerated-content\" href=\"#fs-id1169738238181\">(Figure)<\/a>.<\/p>\n<div id=\"fs-id1169737700320\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} h^{\\prime}(x) & = \\large \\frac{3^x \\ln 3(3^x+2)-3^x \\ln 3(3^x)}{(3^x+2)^2} & & & \\text{Apply the quotient rule.} \\\\ & = \\large \\frac{2 \\cdot 3^x \\ln 3}{(3^x+2)^2} & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738219393\" class=\"textbox examples\">\n<h3>Finding the Slope of a Tangent Line<\/h3>\n<div id=\"fs-id1169738219395\" class=\"exercise\">\n<div id=\"fs-id1169738219397\" class=\"textbox\">\n<p id=\"fs-id1169738219403\">Find the slope of the line tangent to the graph of [latex]y=\\log_2 (3x+1)[\/latex] at [latex]x=1[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738045067\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738045067\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738045067\">To find the slope, we must evaluate [latex]\\frac{dy}{dx}[\/latex] at [latex]x=1[\/latex]. Using <a class=\"autogenerated-content\" href=\"#fs-id1169738186308\">(Figure)<\/a>, we see that<\/p>\n<div id=\"fs-id1169738197861\" class=\"equation unnumbered\">[latex]\\frac{dy}{dx}=\\frac{3}{\\ln 2(3x+1)}[\/latex].<\/div>\n<p id=\"fs-id1169738186987\">By evaluating the derivative at [latex]x=1[\/latex], we see that the tangent line has slope<\/p>\n<div id=\"fs-id1169738187002\" class=\"equation unnumbered\">[latex]\\frac{dy}{dx}|_{x=1} =\\frac{3}{4 \\ln 2}=\\frac{3}{\\ln 16}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738240282\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1169738240285\" class=\"exercise\">\n<div id=\"fs-id1169738240288\" class=\"textbox\">\n<p id=\"fs-id1169738240290\">Find the slope for the line tangent to [latex]y=3^x[\/latex] at [latex]x=2[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737934288\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737934288\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737934288\">[latex]9 \\ln (3)[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1169737934307\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1169737934313\">Evaluate the derivative at [latex]x=2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737933453\" class=\"bc-section section\">\n<h1>Logarithmic Differentiation<\/h1>\n<p id=\"fs-id1169737933458\">At this point, we can take derivatives of functions of the form [latex]y=(g(x))^n[\/latex] for certain values of [latex]n[\/latex], as well as functions of the form [latex]y=b^{g(x)}[\/latex], where [latex]b>0[\/latex] and [latex]b\\ne 1[\/latex]. Unfortunately, we still do not know the derivatives of functions such as [latex]y=x^x[\/latex] or [latex]y=x^{\\pi}[\/latex]. These functions require a technique called<strong> logarithmic differentiation<\/strong>, which allows us to differentiate any function of the form [latex]h(x)=g(x)^{f(x)}[\/latex]. It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of [latex]y=\\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x}[\/latex]. We outline this technique in the following problem-solving strategy.<\/p>\n<div id=\"fs-id1169738197957\" class=\"textbox key-takeaways problem-solving\">\n<h3>Problem-Solving Strategy: Using Logarithmic Differentiation<\/h3>\n<ol id=\"fs-id1169738197963\">\n<li>To differentiate [latex]y=h(x)[\/latex] using logarithmic differentiation, take the natural logarithm of both sides of the equation to obtain [latex]\\ln y=\\ln (h(x))[\/latex].<\/li>\n<li>Use properties of logarithms to expand [latex]\\ln (h(x))[\/latex] as much as possible.<\/li>\n<li>Differentiate both sides of the equation. On the left we will have [latex]\\frac{1}{y}\\frac{dy}{dx}[\/latex].<\/li>\n<li>Multiply both sides of the equation by [latex]y[\/latex] to solve for [latex]\\frac{dy}{dx}[\/latex].<\/li>\n<li>Replace [latex]y[\/latex] by [latex]h(x)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1169738238112\" class=\"textbox examples\">\n<h3>Using Logarithmic Differentiation<\/h3>\n<div id=\"fs-id1169738238115\" class=\"exercise\">\n<div id=\"fs-id1169738238117\" class=\"textbox\">\n<p id=\"fs-id1169738211866\">Find the derivative of [latex]y=(2x^4+1)^{\\tan x}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738211908\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738211908\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738211908\">Use logarithmic differentiation to find this derivative.<\/p>\n<div id=\"fs-id1169738211912\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} \\ln y & = \\ln (2x^4+1)^{\\tan x} & & & \\text{Step 1. Take the natural logarithm of both sides.} \\\\ \\ln y & = \\tan x \\ln (2x^4+1) & & & \\text{Step 2. Expand using properties of logarithms.} \\\\ \\frac{1}{y}\\frac{dy}{dx} & = \\sec^2 x \\ln (2x^4+1)+\\frac{8x^3}{2x^4+1} \\cdot \\tan x & & & \\begin{array}{l}\\text{Step 3. Differentiate both sides. Use the} \\\\ \\text{product rule on the right.} \\end{array} \\\\ \\frac{dy}{dx} & =y \\cdot (\\sec^2 x \\ln (2x^4+1)+\\frac{8x^3}{2x^4+1} \\cdot \\tan x) & & & \\text{Step 4. Multiply by} \\, y \\, \\text{on both sides.} \\\\ \\frac{dy}{dx} & = (2x^4+1)^{\\tan x}(\\sec^2 x \\ln (2x^4+1)+\\frac{8x^3}{2x^4+1} \\cdot \\tan x) & & & \\text{Step 5. Substitute} \\, y=(2x^4+1)^{\\tan x}.\\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738068346\" class=\"textbox examples\">\n<h3>Using Logarithmic Differentiation<\/h3>\n<div id=\"fs-id1169738068348\" class=\"exercise\">\n<div id=\"fs-id1169738068350\" class=\"textbox\">\n<p id=\"fs-id1169738068356\">Find the derivative of [latex]y=\\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x}[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1169738201884\" class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738201884\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738201884\" class=\"hidden-answer\" style=\"display: none\">This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter.<\/p>\n<div id=\"fs-id1169738201891\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} \\ln y & = \\ln \\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x} & & & \\text{Step 1. Take the natural logarithm of both sides.} \\\\ \\ln y & = \\ln x+\\frac{1}{2} \\ln (2x+1)-x \\ln e-3 \\ln \\sin x & & & \\text{Step 2. Expand using properties of logarithms.} \\\\ \\frac{1}{y}\\frac{dy}{dx} & = \\frac{1}{x}+\\frac{1}{2x+1}-1-3\\big(\\frac{\\cos x}{\\sin x}\\big) & & & \\text{Step 3. Differentiate both sides.} \\\\ \\frac{dy}{dx} & = y (\\frac{1}{x}+\\frac{1}{2x+1}-1-3 \\cot x) & & & \\text{Step 4. Multiply by} \\, y \\, \\text{on both sides and simplify.} \\\\ \\frac{dy}{dx} & = \\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x} \\normalsize (\\frac{1}{x}+\\frac{1}{2x+1}-1-3 \\cot x) & & & \\text{Step 5. Substitute} \\, y=\\large \\frac{x\\sqrt{2x+1}}{e^x \\sin^3 x}. \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738228453\" class=\"textbox examples\">\n<h3>Extending the Power Rule<\/h3>\n<div id=\"fs-id1169738228455\" class=\"exercise\">\n<div id=\"fs-id1169738228457\" class=\"textbox\">\n<p id=\"fs-id1169738228463\">Find the derivative of [latex]y=x^r[\/latex] where [latex]r[\/latex] is an arbitrary real number.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738228486\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738228486\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738228486\">The process is the same as in <a class=\"autogenerated-content\" href=\"#fs-id1169738068346\">(Figure)<\/a>, though with fewer complications.<\/p>\n<div id=\"fs-id1169738228494\" class=\"equation unnumbered\">[latex]\\begin{array}{lllll} \\ln y & = \\ln x^r & & & \\text{Step 1. Take the natural logarithm of both sides.} \\\\ \\ln y & = r \\ln x & & & \\text{Step 2. Expand using properties of logarithms.} \\\\ \\frac{1}{y}\\frac{dy}{dx} & = r \\frac{1}{x} & & & \\text{Step 3. Differentiate both sides.} \\\\ \\frac{dy}{dx} & = y \\frac{r}{x} & & & \\text{Step 4. Multiply by} \\, y \\, \\text{on both sides.} \\\\ \\frac{dy}{dx} & = x^r \\frac{r}{x} & & & \\text{Step 5. Substitute} \\, y=x^r. \\\\ \\frac{dy}{dx} & = rx^{r-1} & & & \\text{Simplify.} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737933509\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1169737933513\" class=\"exercise\">\n<div id=\"fs-id1169737933515\" class=\"textbox\">\n<p id=\"fs-id1169737933517\">Use logarithmic differentiation to find the derivative of [latex]y=x^x[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737933537\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737933537\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737933537\">[latex]\\frac{dy}{dx}=x^x(1+\\ln x)[\/latex]<\/p>\n<\/div>\n<div class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1169737933587\">Follow the problem solving strategy.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737933594\" class=\"textbox exercises checkpoint\">\n<div id=\"fs-id1169737933598\" class=\"exercise\">\n<div id=\"fs-id1169737933600\" class=\"textbox\">\n<p id=\"fs-id1169737933602\">Find the derivative of [latex]y=(\\tan x)^{\\pi}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738233543\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738233543\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738233543\">[latex]y^{\\prime}=\\pi (\\tan x)^{\\pi -1} \\sec^2 x[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1169738233590\" class=\"commentary\">\n<h4>Hint<\/h4>\n<p id=\"fs-id1169738233597\">Use the result from <a class=\"autogenerated-content\" href=\"#fs-id1169738228453\">(Figure)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738233608\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id1169738233616\">\n<li>On the basis of the assumption that the exponential function [latex]y=b^x, \\, b>0[\/latex] is continuous everywhere and differentiable at 0, this function is differentiable everywhere and there is a formula for its derivative.<\/li>\n<li>We can use a formula to find the derivative of [latex]y=\\ln x[\/latex], and the relationship [latex]\\log_b x=\\frac{\\ln x}{\\ln b}[\/latex] allows us to extend our differentiation formulas to include logarithms with arbitrary bases.<\/li>\n<li>Logarithmic differentiation allows us to differentiate functions of the form [latex]y=g(x)^{f(x)}[\/latex] or very complex functions by taking the natural logarithm of both sides and exploiting the properties of logarithms before differentiating.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1169738234614\" class=\"key-equations\">\n<h1>Key Equations<\/h1>\n<ul id=\"fs-id1169738234621\">\n<li><strong>Derivative of the natural exponential function<\/strong><br \/>\n[latex]\\frac{d}{dx}(e^{g(x)})=e^{g(x)} g^{\\prime}(x)[\/latex]<\/li>\n<li><strong>Derivative of the natural logarithmic function<\/strong><br \/>\n[latex]\\frac{d}{dx}(\\ln (g(x)))=\\frac{1}{g(x)} g^{\\prime}(x)[\/latex]<\/li>\n<li><strong>Derivative of the general exponential function<\/strong><br \/>\n[latex]\\frac{d}{dx}(b^{g(x)})=b^{g(x)} g^{\\prime}(x) \\ln b[\/latex]<\/li>\n<li><strong>Derivative of the general logarithmic function<\/strong><br \/>\n[latex]\\frac{d}{dx}(\\log_b (g(x)))=\\frac{g^{\\prime}(x)}{g(x) \\ln b}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1169738235113\" class=\"textbox exercises\">\n<p id=\"fs-id1169738235117\">For the following exercises, find [latex]f^{\\prime}(x)[\/latex] for each function.<\/p>\n<div id=\"fs-id1169738235136\" class=\"exercise\">\n<div id=\"fs-id1169738235139\" class=\"textbox\">\n<p id=\"fs-id1169738235141\"><strong>1.\u00a0<\/strong>[latex]f(x)=x^2 e^x[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738235171\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738235171\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738235171\">[latex]f^{\\prime}(x) = 2xe^x+x^2 e^x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738235199\" class=\"exercise\">\n<div id=\"fs-id1169738235202\" class=\"textbox\">\n<p id=\"fs-id1169738235204\"><strong>2.\u00a0<\/strong>[latex]f(x)=\\frac{e^{\u2212x}}{x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738085451\" class=\"exercise\">\n<div id=\"fs-id1169738085453\" class=\"textbox\">\n<p id=\"fs-id1169738085455\"><strong>3.\u00a0<\/strong>[latex]f(x)=e^{x^3 \\ln x}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738085491\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738085491\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738085491\">[latex]f^{\\prime}(x) = e^{x^3 \\ln x}(3x^2 \\ln x+x^2)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738085543\" class=\"exercise\">\n<div id=\"fs-id1169738085545\" class=\"textbox\">\n<p id=\"fs-id1169738085548\"><strong>4.\u00a0<\/strong>[latex]f(x)=\\sqrt{e^{2x}+2x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738240382\" class=\"exercise\">\n<div id=\"fs-id1169738240384\" class=\"textbox\">\n<p id=\"fs-id1169738240386\"><strong>5.\u00a0<\/strong>[latex]f(x)=\\large \\frac{e^x-e^{\u2212x}}{e^x+e^{\u2212x}}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738240440\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738240440\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738240440\">[latex]f^{\\prime}(x) = \\large \\frac{4}{(e^x+e^{\u2212x})^2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738240478\" class=\"exercise\">\n<div id=\"fs-id1169738240480\" class=\"textbox\">\n<p id=\"fs-id1169738240482\"><strong>6.\u00a0<\/strong>[latex]f(x)=\\frac{10^x}{\\ln 10}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738071335\" class=\"exercise\">\n<div id=\"fs-id1169738071337\" class=\"textbox\">\n<p id=\"fs-id1169738071339\"><strong>7.\u00a0<\/strong>[latex]f(x)=2^{4x}+4x^2[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738071376\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738071376\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738071376\">[latex]f^{\\prime}(x) = 2^{4x+2} \\cdot \\ln 2+8x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738071412\" class=\"exercise\">\n<div id=\"fs-id1169738071414\" class=\"textbox\">\n<p id=\"fs-id1169738071416\"><strong>8.\u00a0<\/strong>[latex]f(x)=3^{\\sin 3x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738039846\" class=\"exercise\">\n<div id=\"fs-id1169738039848\" class=\"textbox\">\n<p id=\"fs-id1169738039850\"><strong>9.\u00a0<\/strong>[latex]f(x)=x^{\\pi} \\cdot \\pi^x[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738039882\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738039882\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738039882\">[latex]f^{\\prime}(x) = \\pi x^{\\pi -1} \\cdot \\pi^x + x^{\\pi} \\cdot \\pi^x \\ln \\pi[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738039931\" class=\"exercise\">\n<div id=\"fs-id1169738039933\" class=\"textbox\">\n<p id=\"fs-id1169738039935\"><strong>10.\u00a0<\/strong>[latex]f(x)=\\ln(4x^3+x)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737904496\" class=\"exercise\">\n<div id=\"fs-id1169737904498\" class=\"textbox\">\n<p id=\"fs-id1169737904500\"><strong>11.\u00a0<\/strong>[latex]f(x)=\\ln \\sqrt{5x-7}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737904532\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737904532\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737904532\">[latex]f^{\\prime}(x) = \\frac{5}{2(5x-7)}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737904560\" class=\"exercise\">\n<div id=\"fs-id1169737904562\" class=\"textbox\">\n<p id=\"fs-id1169737904564\"><strong>12.\u00a0<\/strong>[latex]f(x)=x^2 \\ln 9x[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737904626\" class=\"exercise\">\n<div id=\"fs-id1169737904628\" class=\"textbox\">\n<p id=\"fs-id1169737904630\"><strong>13.\u00a0<\/strong>[latex]f(x)=\\log(\\sec x)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737904665\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737904665\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737904665\">[latex]f^{\\prime}(x) = \\frac{\\tan x}{\\ln 10}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738054394\" class=\"exercise\">\n<div id=\"fs-id1169738054396\" class=\"textbox\">\n<p id=\"fs-id1169738054398\"><strong>14.\u00a0<\/strong>[latex]f(x)=\\log_7 (6x^4+3)^5[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738054489\" class=\"exercise\">\n<div id=\"fs-id1169738054491\" class=\"textbox\">\n<p id=\"fs-id1169738054493\"><strong>15.\u00a0<\/strong>[latex]f(x)=2^x \\cdot \\log_3 7^{x^2-4}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738054540\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738054540\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738054540\">[latex]f^{\\prime}(x) = 2^x \\cdot \\ln 2 \\cdot \\log_3 7^{x^2-4} + 2^x \\cdot \\frac{2x \\ln 7}{\\ln 3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169738199886\">For the following exercises, use logarithmic differentiation to find [latex]\\frac{dy}{dx}[\/latex].<\/p>\n<div id=\"fs-id1169738199906\" class=\"exercise\">\n<div id=\"fs-id1169738199908\" class=\"textbox\">\n<p id=\"fs-id1169738199910\"><strong>16.\u00a0<\/strong>[latex]y=x^{\\sqrt{x}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738199977\" class=\"exercise\">\n<div id=\"fs-id1169738199979\" class=\"textbox\">\n<p id=\"fs-id1169738199981\"><strong>17.\u00a0<\/strong>[latex]y=(\\sin 2x)^{4x}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738200016\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738200016\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738200016\">[latex]\\frac{dy}{dx} = (\\sin 2x)^{4x} [4 \\cdot \\ln(\\sin 2x) + 8x \\cdot \\cot 2x][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738092208\" class=\"exercise\">\n<div id=\"fs-id1169738092210\" class=\"textbox\">\n<p id=\"fs-id1169738092212\"><strong>18.\u00a0<\/strong>[latex]y=(\\ln x)^{\\ln x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738092313\" class=\"exercise\">\n<div id=\"fs-id1169738092315\" class=\"textbox\">\n<p id=\"fs-id1169738092318\"><strong>19.\u00a0<\/strong>[latex]y=x^{\\log_2 x}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738092342\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738092342\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738092342\">[latex]\\frac{dy}{dx} = x^{\\log_2 x} \\cdot \\frac{2 \\ln x}{x \\ln 2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738092392\" class=\"exercise\">\n<div id=\"fs-id1169738092394\" class=\"textbox\">\n<p id=\"fs-id1169738092396\"><strong>20.\u00a0<\/strong>[latex]y=(x^2-1)^{\\ln x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738039194\" class=\"exercise\">\n<div id=\"fs-id1169738039196\" class=\"textbox\">\n<p id=\"fs-id1169738039198\"><strong>21.\u00a0<\/strong>[latex]y=x^{\\cot x}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738039222\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738039222\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738039222\">[latex]\\frac{dy}{dx} = x^{\\cot x} \\cdot [\u2212\\csc^2 x \\cdot \\ln x+\\frac{\\cot x}{x}][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738039285\" class=\"exercise\">\n<div id=\"fs-id1169738039287\" class=\"textbox\">\n<p id=\"fs-id1169738093910\"><strong>22.\u00a0<\/strong>[latex]y= \\large \\frac{x+11}{\\sqrt[3]{x^2-4}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738094022\" class=\"exercise\">\n<div id=\"fs-id1169738094024\" class=\"textbox\">\n<p id=\"fs-id1169738094026\"><strong>23.\u00a0<\/strong>[latex]y=x^{-1\/2}(x^2+3)^{2\/3}(3x-4)^4[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738094094\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738094094\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738094094\">[latex]\\frac{dy}{dx} = x^{-1\/2}(x^2+3)^{2\/3}(3x-4)^4 \\cdot [\\frac{-1}{2x}+\\frac{4x}{3(x^2+3)}+\\frac{12}{3x-4}][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738209953\" class=\"exercise\">\n<div id=\"fs-id1169738209955\" class=\"textbox\">\n<p id=\"fs-id1169738209958\"><strong>24. [T]<\/strong> Find an equation of the tangent line to the graph of [latex]f(x)=4xe^{x^2-1}[\/latex] at the point where<\/p>\n<p id=\"fs-id1169738210004\">[latex]x=-1[\/latex]. Graph both the function and the tangent line.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738210057\" class=\"exercise\">\n<div id=\"fs-id1169738210060\" class=\"textbox\">\n<p id=\"fs-id1169738210062\"><strong>25. [T]<\/strong> Find the equation of the line that is normal to the graph of [latex]f(x)=x \\cdot 5^x[\/latex] at the point where [latex]x=1[\/latex]. Graph both the function and the normal line.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738080236\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738080236\" class=\"hidden-answer\" style=\"display: none\"><span id=\"fs-id1169738080243\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205540\/CNX_Calc_Figure_03_09_202.jpg\" alt=\"The function starts at (\u22123, 0), decreases slightly and then increases through the origin and increases to (1.25, 10). There is a straight line marked T(x) with slope \u22121\/(5 + 5 ln 5) and y intercept 5 + 1\/(5 + 5 ln 5).\" \/><\/span><br \/>\n[latex]y=\\frac{-1}{5+5 \\ln 5}x+(5+\\frac{1}{5+5 \\ln 5})[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738080322\" class=\"exercise\">\n<div id=\"fs-id1169738080324\" class=\"textbox\">\n<p id=\"fs-id1169738080326\"><strong>26. [T]<\/strong> Find the equation of the tangent line to the graph of [latex]x^3-x \\ln y+y^3=2x+5[\/latex] at the point where [latex]x=2[\/latex]. (<em>Hint<\/em>: Use implicit differentiation to find [latex]\\frac{dy}{dx}[\/latex].) Graph both the curve and the tangent line.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738094644\" class=\"exercise\">\n<div id=\"fs-id1169738094646\" class=\"textbox\">\n<p id=\"fs-id1169738094648\"><strong>27.\u00a0<\/strong>Consider the function [latex]y=x^{1\/x}[\/latex] for [latex]x>0[\/latex].<\/p>\n<ol id=\"fs-id1169738094680\" style=\"list-style-type: lower-alpha\">\n<li>Determine the points on the graph where the tangent line is horizontal.<\/li>\n<li>Determine the intervals where [latex]y^{\\prime}>0[\/latex] and those where [latex]y^{\\prime}<0[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738094723\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738094723\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738094723\">a. [latex]x=e \\approx 2.718[\/latex]<br \/>\nb. [latex]y^{\\prime}>0[\/latex] on [latex](e,\\infty)[\/latex], and [latex]y^{\\prime}<0[\/latex] on [latex](0,e)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738094772\" class=\"exercise\">\n<div id=\"fs-id1169738094774\" class=\"textbox\">\n<p id=\"fs-id1169738094776\"><strong>28.\u00a0<\/strong>The formula [latex]I(t)=\\frac{\\sin t}{e^t}[\/latex] is the formula for a decaying alternating current.<\/p>\n<ol id=\"fs-id1169738094810\" style=\"list-style-type: lower-alpha\">\n<li>Complete the following table with the appropriate values.<br \/>\n<table id=\"fs-id1169738094824\" class=\"unnumbered\" summary=\"This table has two columns and 10 rows. The first column reads t, 0, \u03c0\/2, \u03c0, 3\u03c0\/2, 2\u03c0, 5\u03c0\/2, 3\u03c0, 7\u03c0\/2, and 4\u03c0. The second column reads (sin t)\/et, (i), (ii), (iii), (iv), (v), (vi), (vii), (viii), and (ix).\">\n<thead>\n<tr valign=\"top\">\n<th>[latex]t[\/latex]<\/th>\n<th>[latex]\\frac{\\sin t}{e^t}[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>0<\/td>\n<td>(i)<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{\\pi}{2}[\/latex]<\/td>\n<td>(ii)<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\pi[\/latex]<\/td>\n<td>(iii)<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{3\\pi}{2}[\/latex]<\/td>\n<td>(iv)<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]2\\pi[\/latex]<\/td>\n<td>(v)<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]2\\pi[\/latex]<\/td>\n<td>(vi)<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]3\\pi[\/latex]<\/td>\n<td>(vii)<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{7\\pi}{2}[\/latex]<\/td>\n<td>(viii)<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]4\\pi[\/latex]<\/td>\n<td>(ix)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>Using only the values in the table, determine where the tangent line to the graph of [latex]I(t)[\/latex] is horizontal.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738229480\" class=\"exercise\">\n<div id=\"fs-id1169738229482\" class=\"textbox\">\n<p id=\"fs-id1169738229484\"><strong>29. [T]<\/strong> The population of Toledo, Ohio, in 2000 was approximately 500,000. Assume the population is increasing at a rate of 5% per year.<\/p>\n<ol id=\"fs-id1169738229493\" style=\"list-style-type: lower-alpha\">\n<li>Write the exponential function that relates the total population as a function of [latex]t[\/latex].<\/li>\n<li>Use a. to determine the rate at which the population is increasing in [latex]t[\/latex] years.<\/li>\n<li>Use b. to determine the rate at which the population is increasing in 10 years.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738229522\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738229522\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738229522\">a. [latex]P=500,000(1.05)^t[\/latex] individuals<br \/>\nb. [latex]P^{\\prime}(t)=24395 \\cdot (1.05)^t[\/latex] individuals per year<br \/>\nc. 39,737 individuals per year<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738229590\" class=\"exercise\">\n<div id=\"fs-id1169738229593\" class=\"textbox\">\n<p id=\"fs-id1169738229595\"><strong>30. [T]<\/strong> An isotope of the element erbium has a half-life of approximately 12 hours. Initially there are 9 grams of the isotope present.<\/p>\n<ol id=\"fs-id1169738229603\" style=\"list-style-type: lower-alpha\">\n<li>Write the exponential function that relates the amount of substance remaining as a function of [latex]t[\/latex], measured in hours.<\/li>\n<li>Use a. to determine the rate at which the substance is decaying in [latex]t[\/latex] hours.<\/li>\n<li>Use b. to determine the rate of decay at [latex]t=4[\/latex] hours.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738220789\" class=\"exercise\">\n<div id=\"fs-id1169738220792\" class=\"textbox\">\n<p id=\"fs-id1169738220794\"><strong>31. [T]<\/strong> The number of cases of influenza in New York City from the beginning of 1960 to the beginning of 1961 is modeled by the function<\/p>\n<p id=\"fs-id1169738220802\">[latex]N(t)=5.3e^{0.093t^2-0.87t}, \\, (0\\le t\\le 4)[\/latex],<\/p>\n<p id=\"fs-id1169738220862\">where [latex]N(t)[\/latex] gives the number of cases (in thousands) and [latex]t[\/latex] is measured in years, with [latex]t=0[\/latex] corresponding to the beginning of 1960.<\/p>\n<ol id=\"fs-id1169738220892\" style=\"list-style-type: lower-alpha\">\n<li>Show work that evaluates [latex]N(0)[\/latex] and [latex]N(4)[\/latex]. Briefly describe what these values indicate about the disease in New York City.<\/li>\n<li>Show work that evaluates [latex]N^{\\prime}(0)[\/latex] and [latex]N^{\\prime}(3)[\/latex]. Briefly describe what these values indicate about the disease in New York City.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169737922881\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169737922881\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737922881\">a. At the beginning of 1960 there were 5.3 thousand cases of the disease in New York City. At the beginning of 1963 there were approximately 723 cases of the disease in New York City.<br \/>\nb. At the beginning of 1960 the number of cases of the disease was decreasing at rate of -4.611 thousand per year; at the beginning of 1963, the number of cases of the disease was decreasing at a rate of -0.2808 thousand per year.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737922901\" class=\"exercise\">\n<div id=\"fs-id1169737922903\" class=\"textbox\">\n<p id=\"fs-id1169737922905\"><strong>32. [T]<\/strong> The <em>relative rate of change<\/em> of a differentiable function [latex]y=f(x)[\/latex] is given by [latex]\\frac{100 \\cdot f^{\\prime}(x)}{f(x)}\\%[\/latex]. One model for population growth is a Gompertz growth function, given by [latex]P(x)=ae^{\u2212b \\cdot e^{\u2212cx}}[\/latex] where [latex]a, \\, b[\/latex], and [latex]c[\/latex] are constants.<\/p>\n<ol id=\"fs-id1169737923026\" style=\"list-style-type: lower-alpha\">\n<li>Find the relative rate of change formula for the generic Gompertz function.<\/li>\n<li>Use a. to find the relative rate of change of a population in [latex]x=20[\/latex] months when [latex]a=204,b=0.0198,[\/latex] and [latex]c=0.15.[\/latex]<\/li>\n<li>Briefly interpret what the result of b. means.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p id=\"fs-id1169738227093\">For the following exercises, use the population of New York City from 1790 to 1860, given in the following table.<\/p>\n<table id=\"fs-id1169738227097\" summary=\"This table has nine rows and two columns. The first row is a header row and it labels each column. The first column header is Years since 1790 and the second column is Population. Under the first column are the values 0, 10, 20, 30, 40, 50, 60, and 70. Under the second column are the values 33,131; 60,515; 96,373; 123,706; 202,300; 312,710; 515,547; and 813,669.\">\n<caption>New York City Population Over Time<br \/>\nSource: http:\/\/en.wikipedia.org\/wiki\/Largest_cities_in_the_United_States_by_population_by_decade<\/caption>\n<\/table>\n<table style=\"border-collapse: collapse;width: 100%\">\n<tbody>\n<tr>\n<td style=\"width: 50%;text-align: center\"><strong>Years since 1790<\/strong><\/td>\n<td style=\"width: 50%;text-align: center\"><strong>Population<\/strong><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;text-align: center\">0<\/td>\n<td style=\"width: 50%;text-align: center\">33,131<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;text-align: center\">10<\/td>\n<td style=\"width: 50%;text-align: center\">60,515<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;text-align: center\">20<\/td>\n<td style=\"width: 50%;text-align: center\">96,373<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;text-align: center\">30<\/td>\n<td style=\"width: 50%;text-align: center\">123,706<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;text-align: center\">40<\/td>\n<td style=\"width: 50%;text-align: center\">202,300<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;text-align: center\">50<\/td>\n<td style=\"width: 50%;text-align: center\">312,710<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;text-align: center\">60<\/td>\n<td style=\"width: 50%;text-align: center\">515,547<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;text-align: center\">70<\/td>\n<td style=\"width: 50%;text-align: center\">813,669<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-id1169738066609\" class=\"exercise\">\n<div id=\"fs-id1169738066611\" class=\"textbox\">\n<p id=\"fs-id1169738066613\"><strong>33. [T]<\/strong> Using a computer program or a calculator, fit a growth curve to the data of the form [latex]p=ab^t[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738066639\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738066639\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738066639\">[latex]p=35741(1.045)^t[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738066665\" class=\"exercise\">\n<div id=\"fs-id1169738066667\" class=\"textbox\">\n<p id=\"fs-id1169738066669\"><strong>34. [T]<\/strong> Using the exponential best fit for the data, write a table containing the derivatives evaluated at each year.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738071156\" class=\"exercise\">\n<div id=\"fs-id1169738071158\" class=\"textbox\">\n<p><strong>35. [T]<\/strong> Using the exponential best fit for the data, write a table containing the second derivatives evaluated at each year.<\/p>\n<\/div>\n<div id=\"fs-id1169738071170\" class=\"textbox shaded\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738071170\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738071170\" class=\"hidden-answer\" style=\"display: none\">\n<table id=\"fs-id1169738071176\" class=\"unnumbered\" summary=\"This table has nine rows and two columns. The first row is a header row and it labels each column. The first column header is Years since 1790 and the second column is P\u2019\u2019. Under the first column are the values 0, 10, 20, 30, 40, 50, 60, and 70. Under the second column are the values 69.25, 107.5, 167.0, 259.4, 402.8, 625.5, 971.4, and 1508.5.\">\n<thead>\n<tr valign=\"top\">\n<th style=\"text-align: center\">Years since 1790<\/th>\n<th style=\"text-align: center\">[latex]P''[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td style=\"text-align: center\">0<\/td>\n<td style=\"text-align: center\">69.25<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center\">10<\/td>\n<td style=\"text-align: center\">107.5<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center\">20<\/td>\n<td style=\"text-align: center\">167.0<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center\">30<\/td>\n<td style=\"text-align: center\">259.4<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center\">40<\/td>\n<td style=\"text-align: center\">402.8<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center\">50<\/td>\n<td style=\"text-align: center\">625.5<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center\">60<\/td>\n<td style=\"text-align: center\">971.4<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"text-align: center\">70<\/td>\n<td style=\"text-align: center\">1508.5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738040664\" class=\"exercise\">\n<div id=\"fs-id1169738040666\" class=\"textbox\">\n<p id=\"fs-id1169738040668\"><strong>36. [T]<\/strong> Using the tables of first and second derivatives and the best fit, answer the following questions:<\/p>\n<ol id=\"fs-id1169738040677\" style=\"list-style-type: lower-alpha\">\n<li>Will the model be accurate in predicting the future population of New York City? Why or why not?<\/li>\n<li>Estimate the population in 2010. Was the prediction correct from a.?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1169738074914\" class=\"definition\">\n<dt>logarithmic differentiation<\/dt>\n<dd id=\"fs-id1169738074919\">is a technique that allows us to differentiate a function by first taking the natural logarithm of both sides of an equation, applying properties of logarithms to simplify the equation, and differentiating implicitly<\/dd>\n<\/dl>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1877\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus I. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/8b89d172-2927-466f-8661-01abc7ccdba4@2.89\">http:\/\/cnx.org\/contents\/8b89d172-2927-466f-8661-01abc7ccdba4@2.89<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/8b89d172-2927-466f-8661-01abc7ccdba4@2.89<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":311,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus I\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/8b89d172-2927-466f-8661-01abc7ccdba4@2.89\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/8b89d172-2927-466f-8661-01abc7ccdba4@2.89\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1877","chapter","type-chapter","status-publish","hentry"],"part":1777,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/wp-json\/pressbooks\/v2\/chapters\/1877","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/wp-json\/pressbooks\/v2\/chapters\/1877\/revisions"}],"predecessor-version":[{"id":2638,"href":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/wp-json\/pressbooks\/v2\/chapters\/1877\/revisions\/2638"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/wp-json\/pressbooks\/v2\/parts\/1777"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/wp-json\/pressbooks\/v2\/chapters\/1877\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/wp-json\/wp\/v2\/media?parent=1877"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=1877"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/wp-json\/wp\/v2\/contributor?post=1877"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-openstax-calculus1\/wp-json\/wp\/v2\/license?post=1877"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}