{"id":383,"date":"2021-02-04T01:20:51","date_gmt":"2021-02-04T01:20:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=383"},"modified":"2022-03-16T05:40:41","modified_gmt":"2022-03-16T05:40:41","slug":"derivative-of-the-logarithmic-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/derivative-of-the-logarithmic-function\/","title":{"raw":"Derivative of the Logarithmic Function","rendered":"Derivative of the Logarithmic Function"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li><span class=\"os-abstract-content\">Find the derivative of logarithmic functions<\/span><\/li>\r\n<\/ul>\r\n<\/div>\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 shaded\">\r\n<h3 style=\"text-align: center;\">The Derivative of the Natural Logarithmic Function<\/h3>\r\n\r\n<hr \/>\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\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{1}{x}[\/latex]<\/div>\r\n&nbsp;\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\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\dfrac{1}{g(x)} g^{\\prime}(x)[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1169738093053\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\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\" style=\"text-align: center;\">[latex]e^y\\frac{dy}{dx}=1[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738216993\">Solving for [latex]\\frac{dy}{dx}[\/latex] yields<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{1}{e^y}[\/latex]<\/div>\r\n&nbsp;\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\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{1}{x}[\/latex]<\/div>\r\n&nbsp;\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\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{1}{f^{\\prime}(g(x))}=\\dfrac{1}{e^{\\ln x}}=\\dfrac{1}{x}[\/latex]<\/div>\r\n&nbsp;\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\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\dfrac{1}{g(x)} g^{\\prime}(x)[\/latex]<\/div>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n&nbsp;\r\n\r\nThe graph of [latex]y=\\ln x[\/latex] and its derivative [latex]\\frac{dy}{dx}=\\frac{1}{x}[\/latex] are shown in Figure 3.\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\/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\" \/> Figure 3.\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]\r\n<div id=\"fs-id1169738211098\" class=\"textbook exercises\">\r\n<h3>Example: Taking a Derivative of a Natural Logarithm<\/h3>\r\n<p id=\"fs-id1169738228377\">Find the derivative of [latex]f(x)=\\ln(x^3+3x-4)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169738221312\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738221312\"]\r\n<p id=\"fs-id1169738221312\">Use the derivative of a natural logarithm directly.<\/p>\r\n\r\n<div id=\"fs-id1169738220266\" class=\"equation unnumbered\" style=\"text-align: center;\">[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]<\/div>\r\n&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738244488\" class=\"textbook exercises\">\r\n<h3>Example: Using Properties of Logarithms in a Derivative<\/h3>\r\n<p id=\"fs-id1169738106144\">Find the derivative of [latex]f(x)=\\ln\\left(\\dfrac{x^2 \\sin x}{2x+1}\\right)[\/latex]<\/p>\r\n[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\" style=\"text-align: center;\">[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]<\/div>\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738219654\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169738219661\">Differentiate: [latex]f(x)=\\ln (3x+2)^5[\/latex].<\/p>\r\n[reveal-answer q=\"644533\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"644533\"]\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[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[\/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=485&amp;end=523&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\/3.9DerivativesOfExponentialAndLogarithmic485to523_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<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 shaded\">\r\n<h3 style=\"text-align: center;\">Derivatives of General Exponential and Logarithmic Functions<\/h3>\r\n\r\n<hr \/>\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\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{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\">\r\n<p style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\dfrac{g^{\\prime}(x)}{g(x) \\ln b}[\/latex]<\/p>\r\n&nbsp;\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\" style=\"text-align: center;\">[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\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=b^{g(x)} g^{\\prime}(x) \\ln b[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"bc-section section\">\r\n<h3>Proof<\/h3>\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\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{1}{x \\ln b}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738211864\">The derivative from above 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\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=y \\ln b=b^x \\ln b[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737145057\">The more general derivative follows from the chain rule.<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n&nbsp;\r\n<div id=\"fs-id1169737145066\" class=\"textbook exercises\">\r\n<h3>Example: Applying Derivative Formulas<\/h3>\r\n<p id=\"fs-id1169738185259\">Find the derivative of [latex]h(x)= \\dfrac{3^x}{3^x+2}[\/latex]<\/p>\r\n[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 the derivative from above.<\/p>\r\n\r\n<div id=\"fs-id1169737700320\" class=\"equation unnumbered\" style=\"text-align: center;\">[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]<\/div>\r\n&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738219393\" class=\"textbook exercises\">\r\n<h3>Example: Finding the Slope of a Tangent Line<\/h3>\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[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]\\dfrac{dy}{dx}[\/latex] at [latex]x=1[\/latex]. Using the derivative above, we see that<\/p>\r\n\r\n<div id=\"fs-id1169738197861\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{dy}{dx}=\\dfrac{3}{\\ln 2(3x+1)}[\/latex]<\/div>\r\n&nbsp;\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\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}|_{x=1} =\\frac{3}{4 \\ln 2}=\\frac{3}{\\ln 16}[\/latex]<\/div>\r\n&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738240282\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\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[reveal-answer q=\"299031\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"299031\"]\r\n<p id=\"fs-id1169737934313\">Evaluate the derivative at [latex]x=2[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[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[\/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=756&amp;end=790&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\/3.9DerivativesOfExponentialAndLogarithmic756to790_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]16162[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li><span class=\"os-abstract-content\">Find the derivative of logarithmic functions<\/span><\/li>\n<\/ul>\n<\/div>\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 shaded\">\n<h3 style=\"text-align: center;\">The Derivative of the Natural Logarithmic Function<\/h3>\n<hr \/>\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\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{1}{x}[\/latex]<\/div>\n<p>&nbsp;<\/p>\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\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\dfrac{1}{g(x)} g^{\\prime}(x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1169738093053\" class=\"bc-section section\">\n<h3>Proof<\/h3>\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\" style=\"text-align: center;\">[latex]e^y\\frac{dy}{dx}=1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738216993\">Solving for [latex]\\frac{dy}{dx}[\/latex] yields<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{1}{e^y}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738219487\">Finally, we substitute [latex]x=e^y[\/latex] to obtain<\/p>\n<div id=\"fs-id1169737145231\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{1}{x}[\/latex]<\/div>\n<p>&nbsp;<\/p>\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\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{1}{f^{\\prime}(g(x))}=\\dfrac{1}{e^{\\ln x}}=\\dfrac{1}{x}[\/latex]<\/div>\n<p>&nbsp;<\/p>\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\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\dfrac{1}{g(x)} g^{\\prime}(x)[\/latex]<\/div>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The graph of [latex]y=\\ln x[\/latex] and its derivative [latex]\\frac{dy}{dx}=\\frac{1}{x}[\/latex] are shown in Figure 3.<\/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\/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\">Figure 3.\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 id=\"fs-id1169738211098\" class=\"textbook exercises\">\n<h3>Example: Taking a Derivative of a Natural Logarithm<\/h3>\n<p id=\"fs-id1169738228377\">Find the derivative of [latex]f(x)=\\ln(x^3+3x-4)[\/latex]<\/p>\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 the derivative of a natural logarithm directly.<\/p>\n<div id=\"fs-id1169738220266\" class=\"equation unnumbered\" style=\"text-align: center;\">[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<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738244488\" class=\"textbook exercises\">\n<h3>Example: Using Properties of Logarithms in a Derivative<\/h3>\n<p id=\"fs-id1169738106144\">Find the derivative of [latex]f(x)=\\ln\\left(\\dfrac{x^2 \\sin x}{2x+1}\\right)[\/latex]<\/p>\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\" style=\"text-align: center;\">[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 class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738219654\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169738219661\">Differentiate: [latex]f(x)=\\ln (3x+2)^5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q644533\">Hint<\/span><\/p>\n<div id=\"q644533\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738073202\">Use a property of logarithms to simplify before taking the derivative.<\/p>\n<\/div>\n<\/div>\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>\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=485&amp;end=523&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\/3.9DerivativesOfExponentialAndLogarithmic485to523_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<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 shaded\">\n<h3 style=\"text-align: center;\">Derivatives of General Exponential and Logarithmic Functions<\/h3>\n<hr \/>\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\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{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\">\n<p style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\dfrac{g^{\\prime}(x)}{g(x) \\ln b}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/li>\n<li>If [latex]y=b^x[\/latex], then\n<div id=\"fs-id1169738224034\" class=\"equation\" style=\"text-align: center;\">[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\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=b^{g(x)} g^{\\prime}(x) \\ln b[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<\/ol>\n<\/div>\n<div class=\"bc-section section\">\n<h3>Proof<\/h3>\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\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\dfrac{1}{x \\ln b}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738211864\">The derivative from above 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\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=y \\ln b=b^x \\ln b[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737145057\">The more general derivative follows from the chain rule.<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1169737145066\" class=\"textbook exercises\">\n<h3>Example: Applying Derivative Formulas<\/h3>\n<p id=\"fs-id1169738185259\">Find the derivative of [latex]h(x)= \\dfrac{3^x}{3^x+2}[\/latex]<\/p>\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 the derivative from above.<\/p>\n<div id=\"fs-id1169737700320\" class=\"equation unnumbered\" style=\"text-align: center;\">[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<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738219393\" class=\"textbook exercises\">\n<h3>Example: Finding the Slope of a Tangent Line<\/h3>\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 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]\\dfrac{dy}{dx}[\/latex] at [latex]x=1[\/latex]. Using the derivative above, we see that<\/p>\n<div id=\"fs-id1169738197861\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{dy}{dx}=\\dfrac{3}{\\ln 2(3x+1)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\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\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}|_{x=1} =\\frac{3}{4 \\ln 2}=\\frac{3}{\\ln 16}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738240282\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q299031\">Hint<\/span><\/p>\n<div id=\"q299031\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169737934313\">Evaluate the derivative at [latex]x=2[\/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-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>\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=756&amp;end=790&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\/3.9DerivativesOfExponentialAndLogarithmic756to790_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=\"ohm16162\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16162&theme=oea&iframe_resize_id=ohm16162&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\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-383\">\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":38,"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 Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-383","chapter","type-chapter","status-publish","hentry"],"part":35,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/383","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":19,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/383\/revisions"}],"predecessor-version":[{"id":4820,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/383\/revisions\/4820"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/35"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/383\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=383"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=383"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=383"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=383"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}