{"id":257,"date":"2023-09-20T22:48:37","date_gmt":"2023-09-20T22:48:37","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/derivatives-of-various-inverse-functions\/"},"modified":"2024-08-05T12:51:24","modified_gmt":"2024-08-05T12:51:24","slug":"derivatives-of-inverse-functions-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/derivatives-of-inverse-functions-learn-it-1\/","title":{"raw":"Derivatives of Inverse Functions: Learn It 1","rendered":"Derivatives of Inverse Functions: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Find the derivative of an inverse function<\/li>\r\n\t<li>Identify the derivatives for inverse trig functions like arcsine, arccosine, and arctangent<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Derivatives of Various Inverse Functions<\/h2>\r\n<p>Understanding the relationship between the derivatives of a function and its inverse is crucial when extending the application of derivatives to inverse functions without needing the limit definition every time. We start by exploring how the derivative of a function and its inverse are interlinked.<\/p>\r\n<p id=\"fs-id1169736613607\">If a function [latex]f(x)[\/latex] is both invertible and differentiable, it logically follows that its inverse [latex]f^{\u22121}(x)[\/latex] should also be differentiable.<\/p>\r\n<p>Consider the function [latex]f(x)[\/latex] and its inverse [latex]f^{\u22121}(x)[\/latex] \u00a0as depicted in Figure 1, where the point on the graph of [latex](a,f ^{\u2212}1 (a))[\/latex] on the graph of [latex]f^{\u22121}(x)[\/latex] and the point [latex](f^{\u22121}(a), a)[\/latex] on the graph of [latex]f(x)[\/latex]\u00a0demonstrate this relationship.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"477\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205433\/CNX_Calc_Figure_03_07_001.jpg\" alt=\"This graph shows a function f(x) and its inverse f\u22121(x). These functions are symmetric about the line y = x. The tangent line of the function f(x) at the point (f\u22121(a), a) and the tangent line of the function f\u22121(x) at (a, f\u22121(a)) are also symmetric about the line y = x. Specifically, if the slope of one were p\/q, then the slope of the other would be q\/p. Lastly, their derivatives are also symmetric about the line y = x.\" width=\"477\" height=\"360\" \/> Figure 1. The tangent lines of a function and its inverse are related; so, too, are the derivatives of these functions.[\/caption]\r\n\r\n<p id=\"fs-id1169739030946\">At these points, the tangent lines of the function and its inverse have reciprocal slopes due to their symmetrical relationship about the line [latex]y=x[\/latex]. Specifically, if the slope of the tangent at [latex]a[\/latex] on [latex]f(x)[\/latex] \u00a0is [latex]f ^{\\prime} (f^{ \u22121} (a))= \\frac{q}{p} \u200b[\/latex], then the slope at [latex]f^{\u22121}(a)[\/latex] on [latex]f^{\u22121}(x)[\/latex] must be [latex]\\frac{q}{p}[\/latex]. This is illustrated in the reciprocal nature of their slopes, confirming that:<\/p>\r\n<div id=\"fs-id1169738913441\" class=\"equation\" style=\"text-align: center;\">[latex](f^{-1})^{\\prime}(a)=\\dfrac{1}{f^{\\prime}(f^{-1}(a))}[\/latex]<\/div>\r\n<p id=\"fs-id1169739028393\">We summarize this result in the following theorem.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">inverse function theorem<\/h3>\r\n<p id=\"fs-id1169736655824\">Let [latex]f(x)[\/latex] be a function that is both invertible and differentiable. Let [latex]y=f^{-1}(x)[\/latex] be the inverse of [latex]f(x)[\/latex]. For all [latex]x[\/latex] satisfying [latex]f^{\\prime}(f^{-1}(x))\\ne 0[\/latex],<\/p>\r\n<div id=\"fs-id1169739040576\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\frac{d}{dx}(f^{-1}(x))=(f^{-1})^{\\prime}(x)=\\dfrac{1}{f^{\\prime}(f^{-1}(x))}[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1169739270778\">Alternatively, if [latex]y=g(x)[\/latex] is the inverse of [latex]f(x)[\/latex], then<\/p>\r\n<div id=\"fs-id1169739010824\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=\\dfrac{1}{f^{\\prime}(g(x))}[\/latex]<\/div>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739007605\">Use the inverse function theorem to find the derivative of [latex]g(x)=\\dfrac{x+2}{x}[\/latex]. Compare the resulting derivative to that obtained by differentiating the function directly.<\/p>\r\n<p>[reveal-answer q=\"fs-id1169738951381\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169738951381\"]<\/p>\r\n<p id=\"fs-id1169738951381\">The inverse of [latex]g(x)=\\frac{x+2}{x}[\/latex] is [latex]f(x)=\\frac{2}{x-1}[\/latex]. Since [latex]g^{\\prime}(x)=\\frac{1}{f^{\\prime}(g(x))}[\/latex], begin by finding [latex]f^{\\prime}(x)[\/latex].<\/p>\r\n<p>Thus,<\/p>\r\n<div id=\"fs-id1169739100123\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{-2}{(x-1)^2}[\/latex] and [latex]f^{\\prime}(g(x))=\\frac{-2}{(g(x)-1)^2}=\\frac{-2}{(\\frac{x+2}{x}-1)^2}=-\\frac{x^2}{2}[\/latex]<\/div>\r\n<p id=\"fs-id1169736655762\">Finally,<\/p>\r\n<p style=\"text-align: center;\">[latex]g^{\\prime}(x)=\\frac{1}{f^{\\prime}(g(x))}=-\\frac{2}{x^2}[\/latex]<\/p>\r\n<p id=\"fs-id1169739000892\">We can verify that this is the correct derivative by applying the quotient rule to [latex]g(x)[\/latex] to obtain<\/p>\r\n<div id=\"fs-id1169739186571\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=-\\frac{2}{x^2}[\/latex]<\/div>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<p><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/0dlA5QJZYsw?controls=0&amp;start=40&amp;end=210&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\r\n<p>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>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.7DerivativesOfInverseFunctions40to210_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.7 Derivatives of Inverse Functions (edited)\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">Find the derivative of [latex]g(x)=\\sqrt[5]{x}[\/latex] by applying the inverse function theorem.\r\n\r\n<p>[reveal-answer q=\"3377621\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"3377621\"]<\/p>\r\n<p id=\"fs-id1169739009099\">Use the fact that [latex]g(x)[\/latex] is the inverse of [latex]f(x)=x^5[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1169739014316\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739014316\"]<\/p>\r\n<p id=\"fs-id1169739014316\">[latex]g(x)=\\frac{1}{5}x^{\u22124\/5}[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p id=\"fs-id1169739326667\">From the previous example, we see that we can use the inverse function theorem to extend the power rule to exponents of the form [latex]\\frac{1}{n}[\/latex], where [latex]n[\/latex] is a positive integer. This extension will ultimately allow us to differentiate [latex]x^q[\/latex], where [latex]q[\/latex] is any rational number.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3 style=\"text-align: left;\">extending the power rule to rational exponents<\/h3>\r\n<p id=\"fs-id1169738960496\">The power rule may be extended to rational exponents. That is, if [latex]n[\/latex] is a positive integer, then<\/p>\r\n<div id=\"fs-id1169736654797\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^{1\/n})=\\frac{1}{n}x^{(1\/n)-1}[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1169736657214\">Also, if [latex]n[\/latex] is a positive integer and [latex]m[\/latex] is an arbitrary integer, then<\/p>\r\n<div id=\"fs-id1169738994019\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^{m\/n})=\\frac{m}{n}x^{(m\/n)-1}[\/latex]<\/div>\r\n<\/section>\r\n<section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n<hr \/>\r\n<p id=\"fs-id1169739000138\">The function [latex]g(x)=x^{1\/n}[\/latex] is the inverse of the function [latex]f(x)=x^n[\/latex]. Since [latex]g^{\\prime}(x)=\\dfrac{1}{f^{\\prime}(g(x))}[\/latex], begin by finding [latex]f^{\\prime}(x)[\/latex]. Thus,<\/p>\r\n<div id=\"fs-id1169736659668\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=nx^{n-1}[\/latex] and [latex]f^{\\prime}(g(x))=n(x^{1\/n})^{n-1}=nx^{(n-1)\/n}[\/latex].<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1169738961739\">Finally,<\/p>\r\n<div id=\"fs-id1169739189161\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=\\dfrac{1}{nx^{(n-1)\/n}}=\\frac{1}{n}x^{(1-n)\/n}=\\frac{1}{n}x^{(1\/n)-1}[\/latex].<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1169736613834\">To differentiate [latex]x^{m\/n}[\/latex] we must rewrite it as [latex](x^{1\/n})^m[\/latex] and apply the chain rule. Thus,<\/p>\r\n<div id=\"fs-id1169739008349\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^{m\/n})=\\frac{d}{dx}((x^{1\/n})^m)=m(x^{1\/n})^{m-1} \\cdot \\frac{1}{n}x^{(1\/n)-1}=\\frac{m}{n}x^{(m\/n)-1}[\/latex].<\/div>\r\n\r\n[latex]_\\blacksquare[\/latex]<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739343740\">Find the equation of the line tangent to the graph of [latex]y=x^{\\frac{2}{3}}[\/latex] at [latex]x=8[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1169739020528\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739020528\"]<\/p>\r\n<p id=\"fs-id1169739020528\">First find [latex]\\frac{dy}{dx}[\/latex] and evaluate it at [latex]x=8[\/latex]. Since<\/p>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\frac{2}{3}x^{-1\/3}[\/latex] and [latex]\\frac{dy}{dx}|_{x=8}=\\frac{1}{3}[\/latex]<\/div>\r\n<p id=\"fs-id1169739234023\">the slope of the tangent line to the graph at [latex]x=8[\/latex] is [latex]\\frac{1}{3}[\/latex].<\/p>\r\n<p id=\"fs-id1169738993930\">Substituting [latex]x=8[\/latex] into the original function, we obtain [latex]y=4[\/latex]. Thus, the tangent line passes through the point [latex](8,4)[\/latex]. Substituting into the point-slope formula for a line and solving for [latex]y[\/latex], we obtain the tangent line<\/p>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\frac{1}{3}x+\\frac{4}{3}[\/latex].<\/div>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/0dlA5QJZYsw?controls=0&amp;start=316&amp;end=425&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>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>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.7DerivativesOfInverseFunctions316to425_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.7 Derivatives of Inverse Functions (edited)\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]33741[\/ohm_question]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Find the derivative of an inverse function<\/li>\n<li>Identify the derivatives for inverse trig functions like arcsine, arccosine, and arctangent<\/li>\n<\/ul>\n<\/section>\n<h2>Derivatives of Various Inverse Functions<\/h2>\n<p>Understanding the relationship between the derivatives of a function and its inverse is crucial when extending the application of derivatives to inverse functions without needing the limit definition every time. We start by exploring how the derivative of a function and its inverse are interlinked.<\/p>\n<p id=\"fs-id1169736613607\">If a function [latex]f(x)[\/latex] is both invertible and differentiable, it logically follows that its inverse [latex]f^{\u22121}(x)[\/latex] should also be differentiable.<\/p>\n<p>Consider the function [latex]f(x)[\/latex] and its inverse [latex]f^{\u22121}(x)[\/latex] \u00a0as depicted in Figure 1, where the point on the graph of [latex](a,f ^{\u2212}1 (a))[\/latex] on the graph of [latex]f^{\u22121}(x)[\/latex] and the point [latex](f^{\u22121}(a), a)[\/latex] on the graph of [latex]f(x)[\/latex]\u00a0demonstrate this relationship.<\/p>\n<figure style=\"width: 477px\" 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\/11205433\/CNX_Calc_Figure_03_07_001.jpg\" alt=\"This graph shows a function f(x) and its inverse f\u22121(x). These functions are symmetric about the line y = x. The tangent line of the function f(x) at the point (f\u22121(a), a) and the tangent line of the function f\u22121(x) at (a, f\u22121(a)) are also symmetric about the line y = x. Specifically, if the slope of one were p\/q, then the slope of the other would be q\/p. Lastly, their derivatives are also symmetric about the line y = x.\" width=\"477\" height=\"360\" \/><figcaption class=\"wp-caption-text\">Figure 1. The tangent lines of a function and its inverse are related; so, too, are the derivatives of these functions.<\/figcaption><\/figure>\n<p id=\"fs-id1169739030946\">At these points, the tangent lines of the function and its inverse have reciprocal slopes due to their symmetrical relationship about the line [latex]y=x[\/latex]. Specifically, if the slope of the tangent at [latex]a[\/latex] on [latex]f(x)[\/latex] \u00a0is [latex]f ^{\\prime} (f^{ \u22121} (a))= \\frac{q}{p} \u200b[\/latex], then the slope at [latex]f^{\u22121}(a)[\/latex] on [latex]f^{\u22121}(x)[\/latex] must be [latex]\\frac{q}{p}[\/latex]. This is illustrated in the reciprocal nature of their slopes, confirming that:<\/p>\n<div id=\"fs-id1169738913441\" class=\"equation\" style=\"text-align: center;\">[latex](f^{-1})^{\\prime}(a)=\\dfrac{1}{f^{\\prime}(f^{-1}(a))}[\/latex]<\/div>\n<p id=\"fs-id1169739028393\">We summarize this result in the following theorem.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">inverse function theorem<\/h3>\n<p id=\"fs-id1169736655824\">Let [latex]f(x)[\/latex] be a function that is both invertible and differentiable. Let [latex]y=f^{-1}(x)[\/latex] be the inverse of [latex]f(x)[\/latex]. For all [latex]x[\/latex] satisfying [latex]f^{\\prime}(f^{-1}(x))\\ne 0[\/latex],<\/p>\n<div id=\"fs-id1169739040576\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\frac{d}{dx}(f^{-1}(x))=(f^{-1})^{\\prime}(x)=\\dfrac{1}{f^{\\prime}(f^{-1}(x))}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739270778\">Alternatively, if [latex]y=g(x)[\/latex] is the inverse of [latex]f(x)[\/latex], then<\/p>\n<div id=\"fs-id1169739010824\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=\\dfrac{1}{f^{\\prime}(g(x))}[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739007605\">Use the inverse function theorem to find the derivative of [latex]g(x)=\\dfrac{x+2}{x}[\/latex]. Compare the resulting derivative to that obtained by differentiating the function directly.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169738951381\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169738951381\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738951381\">The inverse of [latex]g(x)=\\frac{x+2}{x}[\/latex] is [latex]f(x)=\\frac{2}{x-1}[\/latex]. Since [latex]g^{\\prime}(x)=\\frac{1}{f^{\\prime}(g(x))}[\/latex], begin by finding [latex]f^{\\prime}(x)[\/latex].<\/p>\n<p>Thus,<\/p>\n<div id=\"fs-id1169739100123\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{-2}{(x-1)^2}[\/latex] and [latex]f^{\\prime}(g(x))=\\frac{-2}{(g(x)-1)^2}=\\frac{-2}{(\\frac{x+2}{x}-1)^2}=-\\frac{x^2}{2}[\/latex]<\/div>\n<p id=\"fs-id1169736655762\">Finally,<\/p>\n<p style=\"text-align: center;\">[latex]g^{\\prime}(x)=\\frac{1}{f^{\\prime}(g(x))}=-\\frac{2}{x^2}[\/latex]<\/p>\n<p id=\"fs-id1169739000892\">We can verify that this is the correct derivative by applying the quotient rule to [latex]g(x)[\/latex] to obtain<\/p>\n<div id=\"fs-id1169739186571\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=-\\frac{2}{x^2}[\/latex]<\/div>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/0dlA5QJZYsw?controls=0&amp;start=40&amp;end=210&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>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\/3.7DerivativesOfInverseFunctions40to210_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.7 Derivatives of Inverse Functions (edited)&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Find the derivative of [latex]g(x)=\\sqrt[5]{x}[\/latex] by applying the inverse function theorem.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q3377621\">Hint<\/button><\/p>\n<div id=\"q3377621\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739009099\">Use the fact that [latex]g(x)[\/latex] is the inverse of [latex]f(x)=x^5[\/latex].<\/p>\n<\/div>\n<\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739014316\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739014316\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739014316\">[latex]g(x)=\\frac{1}{5}x^{\u22124\/5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1169739326667\">From the previous example, we see that we can use the inverse function theorem to extend the power rule to exponents of the form [latex]\\frac{1}{n}[\/latex], where [latex]n[\/latex] is a positive integer. This extension will ultimately allow us to differentiate [latex]x^q[\/latex], where [latex]q[\/latex] is any rational number.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">extending the power rule to rational exponents<\/h3>\n<p id=\"fs-id1169738960496\">The power rule may be extended to rational exponents. That is, if [latex]n[\/latex] is a positive integer, then<\/p>\n<div id=\"fs-id1169736654797\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^{1\/n})=\\frac{1}{n}x^{(1\/n)-1}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736657214\">Also, if [latex]n[\/latex] is a positive integer and [latex]m[\/latex] is an arbitrary integer, then<\/p>\n<div id=\"fs-id1169738994019\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^{m\/n})=\\frac{m}{n}x^{(m\/n)-1}[\/latex]<\/div>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<p id=\"fs-id1169739000138\">The function [latex]g(x)=x^{1\/n}[\/latex] is the inverse of the function [latex]f(x)=x^n[\/latex]. Since [latex]g^{\\prime}(x)=\\dfrac{1}{f^{\\prime}(g(x))}[\/latex], begin by finding [latex]f^{\\prime}(x)[\/latex]. Thus,<\/p>\n<div id=\"fs-id1169736659668\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=nx^{n-1}[\/latex] and [latex]f^{\\prime}(g(x))=n(x^{1\/n})^{n-1}=nx^{(n-1)\/n}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738961739\">Finally,<\/p>\n<div id=\"fs-id1169739189161\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=\\dfrac{1}{nx^{(n-1)\/n}}=\\frac{1}{n}x^{(1-n)\/n}=\\frac{1}{n}x^{(1\/n)-1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736613834\">To differentiate [latex]x^{m\/n}[\/latex] we must rewrite it as [latex](x^{1\/n})^m[\/latex] and apply the chain rule. Thus,<\/p>\n<div id=\"fs-id1169739008349\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{d}{dx}(x^{m\/n})=\\frac{d}{dx}((x^{1\/n})^m)=m(x^{1\/n})^{m-1} \\cdot \\frac{1}{n}x^{(1\/n)-1}=\\frac{m}{n}x^{(m\/n)-1}[\/latex].<\/div>\n<p>[latex]_\\blacksquare[\/latex]<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739343740\">Find the equation of the line tangent to the graph of [latex]y=x^{\\frac{2}{3}}[\/latex] at [latex]x=8[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739020528\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739020528\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739020528\">First find [latex]\\frac{dy}{dx}[\/latex] and evaluate it at [latex]x=8[\/latex]. Since<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{dy}{dx}=\\frac{2}{3}x^{-1\/3}[\/latex] and [latex]\\frac{dy}{dx}|_{x=8}=\\frac{1}{3}[\/latex]<\/div>\n<p id=\"fs-id1169739234023\">the slope of the tangent line to the graph at [latex]x=8[\/latex] is [latex]\\frac{1}{3}[\/latex].<\/p>\n<p id=\"fs-id1169738993930\">Substituting [latex]x=8[\/latex] into the original function, we obtain [latex]y=4[\/latex]. Thus, the tangent line passes through the point [latex](8,4)[\/latex]. Substituting into the point-slope formula for a line and solving for [latex]y[\/latex], we obtain the tangent line<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\frac{1}{3}x+\\frac{4}{3}[\/latex].<\/div>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/0dlA5QJZYsw?controls=0&amp;start=316&amp;end=425&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>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\/3.7DerivativesOfInverseFunctions316to425_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.7 Derivatives of Inverse Functions (edited)&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm33741\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=33741&theme=lumen&iframe_resize_id=ohm33741&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":15,"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\":\"3.7 Derivatives of Inverse Functions (edited)\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":560,"module-header":"learn_it","content_attributions":[{"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":"3.7 Derivatives of Inverse Functions (edited)","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/257"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":9,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/257\/revisions"}],"predecessor-version":[{"id":4527,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/257\/revisions\/4527"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/560"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/257\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=257"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=257"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=257"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=257"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}