{"id":382,"date":"2021-02-04T01:20:17","date_gmt":"2021-02-04T01:20:17","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=382"},"modified":"2022-03-16T05:40:07","modified_gmt":"2022-03-16T05:40:07","slug":"derivative-of-the-exponential-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/derivative-of-the-exponential-function\/","title":{"raw":"Derivative of the Exponential Function","rendered":"Derivative of the Exponential Function"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Find the derivative of exponential functions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Derivative of the Exponential Function<\/h2>\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\" style=\"text-align: center;\">[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&nbsp;\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>. Figure 1 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[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\" \/> Figure 1. The graph of [latex]E(x)=e^x[\/latex] is between [latex]y=2^x[\/latex] and [latex]y=3^x[\/latex].[\/caption]\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\" style=\"text-align: center;\">[latex]B^{\\prime}(0)=\\underset{x\\to 0}{\\lim}\\dfrac{b^x-b^0}{x-0}=\\underset{x\\to 0}{\\lim}\\dfrac{b^x-1}{x} \\approx \\dfrac{b^x-1}{x}[\/latex]<\/div>\r\n&nbsp;\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\" style=\"text-align: center;\">[latex]\\dfrac{b^{-0.00001}-1}{-0.00001}&lt;B^{\\prime}(0)&lt;\\dfrac{b^{0.00001}-1}{0.00001}[\/latex]<\/div>\r\n&nbsp;\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 Figure 2. This line is tangent to the graph of [latex]E(x)=e^x[\/latex] at [latex]x=0[\/latex].<\/p>\r\n&nbsp;\r\n\r\n[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\" \/> Figure 2. The tangent line to [latex]E(x)=e^x[\/latex] at [latex]x=0[\/latex] has slope 1.[\/caption]\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\" style=\"text-align: center;\">[latex]B^{\\prime}(0)=\\underset{h\\to 0}{\\lim}\\dfrac{b^{0+h}-b^0}{h}=\\underset{h\\to 0}{\\lim}\\dfrac{b^h-1}{h}[\/latex]<\/div>\r\n&nbsp;\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\" style=\"text-align: center;\">[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&nbsp;\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\" style=\"text-align: center;\">[latex]B^{\\prime}(x)=b^x B^{\\prime}(0)[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<div class=\"equation\" style=\"text-align: left;\">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.)<\/div>\r\n<div id=\"fs-id1169738226753\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Derivative of the Natural Exponential Function<\/h3>\r\n\r\n<hr \/>\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\" style=\"text-align: center;\">[latex]E^{\\prime}(x)=e^x[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737142141\">In general,<\/p>\r\n\r\n<div id=\"fs-id1169738124964\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(e^{g(x)})=e^{g(x)} g^{\\prime}(x)[\/latex]<\/div>\r\n<div><\/div>\r\n<\/div>\r\nIf it helps, think of the formula as the chain rule being applied to natural exponential functions. The derivative of [latex]{e}[\/latex] raised to the power of a function will simply be\u00a0[latex]{e}[\/latex] raised to the power of the function multiplied by the derivative of that function.\r\n<div id=\"fs-id1169738223456\" class=\"textbook exercises\">\r\n<h3>Example: Derivative of an Exponential Function<\/h3>\r\n<p id=\"fs-id1169738187154\">Find the derivative of [latex]f(x)=e^{\\tan (2x)}[\/latex].<\/p>\r\n[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\" style=\"text-align: center;\">[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]<\/div>\r\n&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737140879\" class=\"textbook exercises\">\r\n<h3>Example: Combining Differentiation Rules<\/h3>\r\n<p id=\"fs-id1169737766547\">Find the derivative of [latex]y=\\dfrac{e^{x^2}}{x}[\/latex].<\/p>\r\n[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\" style=\"text-align: center;\">[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]<\/div>\r\n&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738152550\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169737923851\">Find the derivative of [latex]h(x)=xe^{2x}[\/latex].<\/p>\r\n[reveal-answer q=\"67723309\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"67723309\"]\r\n<p id=\"fs-id1169738215078\">Don\u2019t forget to use the product rule.<\/p>\r\n[\/hidden-answer]\r\n\r\n[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[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_nzxTKiFPpo?controls=0&amp;start=247&amp;end=287&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.9DerivativesOfExponentialAndLogarithmic247to287_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.9 Derivatives of Exponential and Logarithmic Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]33753[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169737949367\" class=\"textbook exercises\">\r\n<h3>Example: Applying the Natural Exponential Function<\/h3>\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[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\" style=\"text-align: center;\">[latex]\\large \\frac{A^{\\prime}(t)}{A(t)} \\normalsize = \\large \\frac{300e^{0.3t}}{1000e^{0.3t}}=0.3[\/latex]<\/div>\r\n&nbsp;\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\nWatch the following video to see the worked solution to Example: Applying the Natural Exponential Function.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_nzxTKiFPpo?controls=0&amp;start=163&amp;end=240&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.9DerivativesOfExponentialAndLogarithmic163to240_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.9 Derivatives of Exponential and Logarithmic Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1169738186819\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\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[reveal-answer q=\"883902\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"883902\"]\r\n<p id=\"fs-id1169738225603\">Find [latex]A^{\\prime}(4)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[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[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find the derivative of exponential functions<\/li>\n<\/ul>\n<\/div>\n<h2>Derivative of the Exponential Function<\/h2>\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\" style=\"text-align: center;\">[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>&nbsp;<\/p>\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>. Figure 1 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 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\">Figure 1. 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<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\" style=\"text-align: center;\">[latex]B^{\\prime}(0)=\\underset{x\\to 0}{\\lim}\\dfrac{b^x-b^0}{x-0}=\\underset{x\\to 0}{\\lim}\\dfrac{b^x-1}{x} \\approx \\dfrac{b^x-1}{x}[\/latex]<\/div>\n<p>&nbsp;<\/p>\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\" style=\"text-align: center;\">[latex]\\dfrac{b^{-0.00001}-1}{-0.00001}<B^{\\prime}(0)<\\dfrac{b^{0.00001}-1}{0.00001}[\/latex]<\/div>\n<p>&nbsp;<\/p>\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 Figure 2. This line is tangent to the graph of [latex]E(x)=e^x[\/latex] at [latex]x=0[\/latex].<\/p>\n<p>&nbsp;<\/p>\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\">Figure 2. The tangent line to [latex]E(x)=e^x[\/latex] at [latex]x=0[\/latex] has slope 1.<\/p>\n<\/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\" style=\"text-align: center;\">[latex]B^{\\prime}(0)=\\underset{h\\to 0}{\\lim}\\dfrac{b^{0+h}-b^0}{h}=\\underset{h\\to 0}{\\lim}\\dfrac{b^h-1}{h}[\/latex]<\/div>\n<p>&nbsp;<\/p>\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\" style=\"text-align: center;\">[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>&nbsp;<\/p>\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\" style=\"text-align: center;\">[latex]B^{\\prime}(x)=b^x B^{\\prime}(0)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<div class=\"equation\" style=\"text-align: left;\">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.)<\/div>\n<div id=\"fs-id1169738226753\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Derivative of the Natural Exponential Function<\/h3>\n<hr \/>\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\" style=\"text-align: center;\">[latex]E^{\\prime}(x)=e^x[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737142141\">In general,<\/p>\n<div id=\"fs-id1169738124964\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(e^{g(x)})=e^{g(x)} g^{\\prime}(x)[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<p>If it helps, think of the formula as the chain rule being applied to natural exponential functions. The derivative of [latex]{e}[\/latex] raised to the power of a function will simply be\u00a0[latex]{e}[\/latex] raised to the power of the function multiplied by the derivative of that function.<\/p>\n<div id=\"fs-id1169738223456\" class=\"textbook exercises\">\n<h3>Example: Derivative of an Exponential Function<\/h3>\n<p id=\"fs-id1169738187154\">Find the derivative of [latex]f(x)=e^{\\tan (2x)}[\/latex].<\/p>\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\" style=\"text-align: center;\">[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<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737140879\" class=\"textbook exercises\">\n<h3>Example: Combining Differentiation Rules<\/h3>\n<p id=\"fs-id1169737766547\">Find the derivative of [latex]y=\\dfrac{e^{x^2}}{x}[\/latex].<\/p>\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\" style=\"text-align: center;\">[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<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738152550\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169737923851\">Find the derivative of [latex]h(x)=xe^{2x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q67723309\">Hint<\/span><\/p>\n<div id=\"q67723309\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738215078\">Don\u2019t forget to use the product rule.<\/p>\n<\/div>\n<\/div>\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>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_nzxTKiFPpo?controls=0&amp;start=247&amp;end=287&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.9DerivativesOfExponentialAndLogarithmic247to287_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.9 Derivatives of Exponential and Logarithmic Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm33753\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=33753&theme=oea&iframe_resize_id=ohm33753&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1169737949367\" class=\"textbook exercises\">\n<h3>Example: Applying the Natural Exponential Function<\/h3>\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 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\" style=\"text-align: center;\">[latex]\\large \\frac{A^{\\prime}(t)}{A(t)} \\normalsize = \\large \\frac{300e^{0.3t}}{1000e^{0.3t}}=0.3[\/latex]<\/div>\n<p>&nbsp;<\/p>\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<p>Watch the following video to see the worked solution to Example: Applying the Natural Exponential Function.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_nzxTKiFPpo?controls=0&amp;start=163&amp;end=240&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.9DerivativesOfExponentialAndLogarithmic163to240_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.9 Derivatives of Exponential and Logarithmic Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1169738186819\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q883902\">Hint<\/span><\/p>\n<div id=\"q883902\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738225603\">Find [latex]A^{\\prime}(4)[\/latex].<\/p>\n<\/div>\n<\/div>\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>\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-382\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>3.9 Derivatives of Exponential and Logarithmic Functions. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/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>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/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":17533,"menu_order":37,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"3.9 Derivatives of Exponential and Logarithmic Functions\",\"author\":\"Ryan 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