{"id":377,"date":"2021-02-04T01:19:19","date_gmt":"2021-02-04T01:19:19","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=377"},"modified":"2022-03-16T05:35:47","modified_gmt":"2022-03-16T05:35:47","slug":"derivatives-of-various-inverse-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/derivatives-of-various-inverse-functions\/","title":{"raw":"Derivatives of Various Inverse Functions","rendered":"Derivatives of Various Inverse Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate the derivative of an inverse function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1169738951769\" class=\"bc-section section\">\r\n<p id=\"fs-id1169736613607\">We begin by considering a function and its inverse. If [latex]f(x)[\/latex] is both invertible and differentiable, it seems reasonable that the inverse of [latex]f(x)[\/latex] is also differentiable. Figure 1 shows the relationship between a function [latex]f(x)[\/latex] and its inverse [latex]f^{-1}(x)[\/latex]. Look at the point [latex](a,f^{-1}(a))[\/latex] on the graph of [latex]f^{-1}(x)[\/latex] having a tangent line with a slope of [latex](f^{-1})^{\\prime}(a)=\\frac{p}{q}[\/latex]. This point corresponds to a point [latex](f^{-1}(a),a)[\/latex] on the graph of [latex]f(x)[\/latex] having a tangent line with a slope of [latex]f^{\\prime}(f^{-1}(a))=\\frac{q}{p}[\/latex]. Thus, if [latex]f^{-1}(x)[\/latex] is differentiable at [latex]a[\/latex], then it must be the case that<\/p>\r\n\r\n<div id=\"fs-id1169738950018\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](f^{-1})^{\\prime}(a)=\\dfrac{1}{f^{\\prime}(f^{-1}(a))}[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<div>\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<\/div>\r\n<p id=\"fs-id1169739030946\">We may also derive the formula for the derivative of the inverse by first recalling that [latex]x=f(f^{-1}(x))[\/latex]. Then by differentiating both sides of this equation (using the chain rule on the right), we obtain<\/p>\r\n\r\n<div id=\"fs-id1169738877985\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1=f^{\\prime}(f^{-1}(x))(f^{-1})^{\\prime}(x))[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739172653\">Solving for [latex](f^{-1})^{\\prime}(x)[\/latex], we obtain<\/p>\r\n\r\n<div id=\"fs-id1169738913441\" class=\"equation\" style=\"text-align: center;\">[latex](f^{-1})^{\\prime}(x)=\\dfrac{1}{f^{\\prime}(f^{-1}(x))}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739028393\">We summarize this result in the following theorem.<\/p>\r\n\r\n<div id=\"fs-id1169739019409\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Inverse Function Theorem<\/h3>\r\n\r\n<hr \/>\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\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&nbsp;\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\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&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1169739001941\" class=\"textbook exercises\">\r\n<h3>Example: Applying the Inverse Function Theorem, 1<\/h3>\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[reveal-answer q=\"fs-id1169738951381\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738951381\"]\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]. Thus,<\/p>\r\n\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&nbsp;\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\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&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Applying the Inverse Function Theorem, 1.\r\n\r\n<center><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><\/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\/3.7DerivativesOfInverseFunctions40to210_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>.[\/hidden-answer]\r\n<div id=\"fs-id1169739212836\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169739269403\">Use the inverse function theorem to find the derivative of [latex]g(x)=\\dfrac{1}{x+2}.[\/latex] Compare the result obtained by differentiating [latex]g(x)[\/latex] directly.<\/p>\r\n[reveal-answer q=\"336869\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"336869\"]\r\n[latex]g^{\\prime}(x)=-\\frac{1}{(x+2)^2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<strong>Hint<\/strong>\r\n\r\nUse the preceding example as a guide.\r\n\r\n<\/div>\r\n<div id=\"fs-id1169738977168\" class=\"textbook exercises\">\r\n<h3>Example: Applying the Inverse Function Theorem, 2<\/h3>\r\n<p id=\"fs-id1169739036425\">Use the inverse function theorem to find the derivative of [latex]g(x)=\\sqrt[3]{x}[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169738969956\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738969956\"]\r\n<p id=\"fs-id1169738969956\">The function [latex]g(x)=\\sqrt[3]{x}[\/latex] is the inverse of the function [latex]f(x)=x^3[\/latex]. Since [latex]g^{\\prime}(x)=\\frac{1}{f^{\\prime}(g(x))}[\/latex], begin by finding [latex]f^{\\prime}(x)[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1169739305067\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=3x^2[\/latex] and [latex]f^{\\prime}(g(x))=3(\\sqrt[3]{x})^2=3x^{2\/3}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739034112\">Finally,<\/p>\r\n\r\n<div id=\"fs-id1169739340252\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=\\frac{1}{3x^{2\/3}}=\\frac{1}{3}x^{-2\/3}[\/latex]<\/div>\r\n&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169739044633\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169738835681\">Find the derivative of [latex]g(x)=\\sqrt[5]{x}[\/latex] by applying the inverse function theorem.<\/p>\r\n[reveal-answer q=\"3377621\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3377621\"]\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[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739014316\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739014316\"]\r\n<p id=\"fs-id1169739014316\">[latex]g(x)=\\frac{1}{5}x^{\u22124\/5}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\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\r\n<div id=\"fs-id1169736611442\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Extending the Power Rule to Rational Exponents<\/h3>\r\n\r\n<hr \/>\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\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&nbsp;\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\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&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-id1169739269836\" class=\"bc-section section\">\r\n<h3>Proof<\/h3>\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\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&nbsp;\r\n<p id=\"fs-id1169738961739\">Finally,<\/p>\r\n\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&nbsp;\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\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<div class=\"equation unnumbered\" style=\"text-align: left;\">[latex]_\\blacksquare[\/latex]<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1169739269836\" class=\"bc-section section\">\r\n<div id=\"fs-id1169739298630\" class=\"textbook exercises\">\r\n<h3>Example: Applying the Power Rule to a Rational Power<\/h3>\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[reveal-answer q=\"fs-id1169739020528\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739020528\"]\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\r\n<div class=\"equation unnumbered\">[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&nbsp;\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\r\n<div class=\"equation unnumbered\">[latex]y=\\frac{1}{3}x+\\frac{4}{3}[\/latex].[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Applying the Power Rule to a Rational Power.\r\n\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\"><\/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\/3.7DerivativesOfInverseFunctions316to425_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>.[\/hidden-answer]\r\n<div id=\"fs-id1169736654814\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169736607624\">Find the derivative of [latex]s(t)=\\sqrt{2t+1}[\/latex].<\/p>\r\n[reveal-answer q=\"8665521\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"8665521\"]\r\n<p id=\"fs-id1169736596062\">Rewrite as [latex]s(t)=(2t+1)^{1\/2}[\/latex] and use the chain rule.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739183503\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739183503\"]\r\n<p id=\"fs-id1169739183503\">[latex]s^{\\prime}(t)=(2t+1)^{\u22121\/2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169737795671\" class=\"bc-section section\"><\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]33741[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate the derivative of an inverse function<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1169738951769\" class=\"bc-section section\">\n<p id=\"fs-id1169736613607\">We begin by considering a function and its inverse. If [latex]f(x)[\/latex] is both invertible and differentiable, it seems reasonable that the inverse of [latex]f(x)[\/latex] is also differentiable. Figure 1 shows the relationship between a function [latex]f(x)[\/latex] and its inverse [latex]f^{-1}(x)[\/latex]. Look at the point [latex](a,f^{-1}(a))[\/latex] on the graph of [latex]f^{-1}(x)[\/latex] having a tangent line with a slope of [latex](f^{-1})^{\\prime}(a)=\\frac{p}{q}[\/latex]. This point corresponds to a point [latex](f^{-1}(a),a)[\/latex] on the graph of [latex]f(x)[\/latex] having a tangent line with a slope of [latex]f^{\\prime}(f^{-1}(a))=\\frac{q}{p}[\/latex]. Thus, if [latex]f^{-1}(x)[\/latex] is differentiable at [latex]a[\/latex], then it must be the case that<\/p>\n<div id=\"fs-id1169738950018\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](f^{-1})^{\\prime}(a)=\\dfrac{1}{f^{\\prime}(f^{-1}(a))}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<div>\n<div style=\"width: 487px\" 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\" \/><\/p>\n<p 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.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1169739030946\">We may also derive the formula for the derivative of the inverse by first recalling that [latex]x=f(f^{-1}(x))[\/latex]. Then by differentiating both sides of this equation (using the chain rule on the right), we obtain<\/p>\n<div id=\"fs-id1169738877985\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]1=f^{\\prime}(f^{-1}(x))(f^{-1})^{\\prime}(x))[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739172653\">Solving for [latex](f^{-1})^{\\prime}(x)[\/latex], we obtain<\/p>\n<div id=\"fs-id1169738913441\" class=\"equation\" style=\"text-align: center;\">[latex](f^{-1})^{\\prime}(x)=\\dfrac{1}{f^{\\prime}(f^{-1}(x))}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739028393\">We summarize this result in the following theorem.<\/p>\n<div id=\"fs-id1169739019409\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Inverse Function Theorem<\/h3>\n<hr \/>\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<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1169739001941\" class=\"textbook exercises\">\n<h3>Example: Applying the Inverse Function Theorem, 1<\/h3>\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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738951381\">Show Solution<\/span><\/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]. 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>&nbsp;<\/p>\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>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Applying the Inverse Function Theorem, 1.<\/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=40&amp;end=210&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\/3.7DerivativesOfInverseFunctions40to210_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>.<\/div>\n<\/div>\n<div id=\"fs-id1169739212836\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169739269403\">Use the inverse function theorem to find the derivative of [latex]g(x)=\\dfrac{1}{x+2}.[\/latex] Compare the result obtained by differentiating [latex]g(x)[\/latex] directly.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q336869\">Show Solution<\/span><\/p>\n<div id=\"q336869\" class=\"hidden-answer\" style=\"display: none\">\n[latex]g^{\\prime}(x)=-\\frac{1}{(x+2)^2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><strong>Hint<\/strong><\/p>\n<p>Use the preceding example as a guide.<\/p>\n<\/div>\n<div id=\"fs-id1169738977168\" class=\"textbook exercises\">\n<h3>Example: Applying the Inverse Function Theorem, 2<\/h3>\n<p id=\"fs-id1169739036425\">Use the inverse function theorem to find the derivative of [latex]g(x)=\\sqrt[3]{x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738969956\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169738969956\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738969956\">The function [latex]g(x)=\\sqrt[3]{x}[\/latex] is the inverse of the function [latex]f(x)=x^3[\/latex]. Since [latex]g^{\\prime}(x)=\\frac{1}{f^{\\prime}(g(x))}[\/latex], begin by finding [latex]f^{\\prime}(x)[\/latex]. Thus,<\/p>\n<div id=\"fs-id1169739305067\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=3x^2[\/latex] and [latex]f^{\\prime}(g(x))=3(\\sqrt[3]{x})^2=3x^{2\/3}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739034112\">Finally,<\/p>\n<div id=\"fs-id1169739340252\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=\\frac{1}{3x^{2\/3}}=\\frac{1}{3}x^{-2\/3}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169739044633\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169738835681\">Find the derivative of [latex]g(x)=\\sqrt[5]{x}[\/latex] by applying the inverse function theorem.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3377621\">Hint<\/span><\/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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739014316\">Show Solution<\/span><\/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<\/div>\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<div id=\"fs-id1169736611442\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Extending the Power Rule to Rational Exponents<\/h3>\n<hr \/>\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<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-id1169739269836\" class=\"bc-section section\">\n<h3>Proof<\/h3>\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<div class=\"equation unnumbered\" style=\"text-align: left;\">[latex]_\\blacksquare[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<div><\/div>\n<div><\/div>\n<div id=\"fs-id1169739269836\" class=\"bc-section section\">\n<div id=\"fs-id1169739298630\" class=\"textbook exercises\">\n<h3>Example: Applying the Power Rule to a Rational Power<\/h3>\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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739020528\">Show Solution<\/span><\/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\">[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>&nbsp;<\/p>\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\">[latex]y=\\frac{1}{3}x+\\frac{4}{3}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Applying the Power Rule to a Rational Power.<\/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\"><\/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\/3.7DerivativesOfInverseFunctions316to425_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>.<\/div>\n<\/div>\n<div id=\"fs-id1169736654814\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169736607624\">Find the derivative of [latex]s(t)=\\sqrt{2t+1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q8665521\">Hint<\/span><\/p>\n<div id=\"q8665521\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736596062\">Rewrite as [latex]s(t)=(2t+1)^{1\/2}[\/latex] and use the chain 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-id1169739183503\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739183503\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739183503\">[latex]s^{\\prime}(t)=(2t+1)^{\u22121\/2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737795671\" class=\"bc-section section\"><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm33741\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=33741&theme=oea&iframe_resize_id=ohm33741&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-377\">\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>3.7 Derivatives of Inverse Functions (edited). <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":30,"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-377","chapter","type-chapter","status-publish","hentry"],"part":35,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/377","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":24,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/377\/revisions"}],"predecessor-version":[{"id":4816,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/377\/revisions\/4816"}],"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\/377\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=377"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=377"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=377"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=377"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}