{"id":378,"date":"2021-02-04T01:19:38","date_gmt":"2021-02-04T01:19:38","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=378"},"modified":"2022-03-16T05:36:23","modified_gmt":"2022-03-16T05:36:23","slug":"derivatives-of-inverse-trigonometric-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/derivatives-of-inverse-trigonometric-functions\/","title":{"raw":"Derivatives of Inverse Trigonometric Functions","rendered":"Derivatives of Inverse Trigonometric Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Recognize the derivatives of the standard inverse trigonometric functions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1169739301198\" class=\"bc-section section\">\r\n<p id=\"fs-id1169739029363\">We now turn our attention to finding derivatives of inverse trigonometric functions. These derivatives will prove invaluable in the study of integration later in this text. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are actually algebraic functions. Previously, derivatives of algebraic functions have proven to be algebraic functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Here, for the first time, we see that the derivative of a function need not be of the same type as the original function.<\/p>\r\n\r\n<div id=\"fs-id1169739208944\" class=\"textbook exercises\">\r\n<h3>Example: Derivative of the Inverse Sine Function<\/h3>\r\n<p id=\"fs-id1169739188589\">Use the inverse function theorem to find the derivative of [latex]g(x)=\\sin^{-1} x[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169739269789\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739269789\"]\r\n<p id=\"fs-id1169739269789\">Since for [latex]x[\/latex] in the interval [latex][-\\frac{\\pi}{2},\\frac{\\pi}{2}], \\, f(x)= \\sin x[\/latex] is the inverse of [latex]g(x)= \\sin^{-1} x[\/latex], begin by finding [latex]f^{\\prime}(x)[\/latex]. Since<\/p>\r\n\r\n<div id=\"fs-id1169736611678\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)= \\cos x[\/latex] and [latex]f^{\\prime}(g(x))= \\cos (\\sin^{-1} x)=\\sqrt{1-x^2}[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739347090\">we see that<\/p>\r\n\r\n<div id=\"fs-id1169739369239\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=\\frac{d}{dx}(\\sin^{-1} x)=\\frac{1}{f^{\\prime}(g(x))}=\\frac{1}{\\sqrt{1-x^2}}[\/latex]<\/div>\r\n&nbsp;\r\n<h4>Analysis<\/h4>\r\n<p id=\"fs-id1169739111057\">To see that [latex]\\cos (\\sin^{-1} x)=\\sqrt{1-x^2}[\/latex], consider the following argument. Set [latex]\\sin^{-1} x=\\theta[\/latex]. In this case, [latex]\\sin \\theta =x[\/latex] where [latex]-\\frac{\\pi}{2}\\le \\theta \\le \\frac{\\pi}{2}[\/latex]. We begin by considering the case where [latex]0&lt;\\theta &lt;\\frac{\\pi}{2}[\/latex]. Since [latex]\\theta[\/latex] is an acute angle, we may construct a right triangle having acute angle [latex]\\theta[\/latex], a hypotenuse of length 1, and the side opposite angle [latex]\\theta [\/latex] having length [latex]x[\/latex]. From the Pythagorean theorem, the side adjacent to angle [latex]\\theta[\/latex] has length [latex]\\sqrt{1-x^2}[\/latex]. This triangle is shown in Figure 2. Using the triangle, we see that [latex]\\cos (\\sin^{-1} x)= \\cos \\theta =\\sqrt{1-x^2}[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"426\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205436\/CNX_Calc_Figure_03_07_002.jpg\" alt=\"A right triangle with angle \u03b8, opposite side x, hypotenuse 1, and adjacent side equal to the square root of the quantity (1 \u2013 x2).\" width=\"426\" height=\"205\" \/> Figure 2. Using a right triangle having acute angle [latex]\\theta[\/latex], a hypotenuse of length 1, and the side opposite angle [latex]\\theta[\/latex] having length [latex]x[\/latex], we can see that [latex]\\cos (\\sin^{-1} x)= \\cos \\theta =\\sqrt{1-x^2}[\/latex].[\/caption]\r\n<div><\/div>\r\n<div id=\"fs-id1169739303300\" class=\"equation unnumbered\">In the case where [latex]-\\frac{\\pi}{2}&lt;\\theta &lt;0[\/latex], we make the observation that [latex]0&lt;-\\theta&lt;\\frac{\\pi}{2}[\/latex] and hence [latex]\\cos (\\sin^{-1} x)= \\cos \\theta = \\cos (\u2212\\theta )=\\sqrt{1-x^2}[\/latex].<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1169736611289\">Now if [latex]\\theta =\\frac{\\pi}{2}[\/latex] or [latex]\\theta =-\\frac{\\pi}{2}, \\, x=1[\/latex] or [latex]x=-1[\/latex], and since in either case [latex]\\cos \\theta =0[\/latex] and [latex]\\sqrt{1-x^2}=0[\/latex], we have<\/p>\r\n\r\n<div id=\"fs-id1169739340290\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\cos (\\sin^{-1} x)= \\cos \\theta =\\sqrt{1-x^2}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739190632\">Consequently, in all cases, [latex]\\cos (\\sin^{-1} x)=\\sqrt{1-x^2}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169736662939\" class=\"textbook exercises\">\r\n<h3>Example: Applying the Chain Rule to the Inverse Sine Function<\/h3>\r\n<p id=\"fs-id1169739190553\">Apply the chain rule to the formula derived in <em>Example: Applying the Inverse Function Theorem<\/em>\r\nto find the derivative of [latex]h(x)=\\sin^{-1} (g(x))[\/latex] and use this result to find the derivative of [latex]h(x)=\\sin^{-1}(2x^3)[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169739189899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739189899\"]\r\n<p id=\"fs-id1169739189899\">Applying the chain rule to [latex]h(x)=\\sin^{-1} (g(x))[\/latex], we have<\/p>\r\n\r\n<div id=\"fs-id1169736596017\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\dfrac{1}{\\sqrt{1-(g(x))^2}}g^{\\prime}(x)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739269950\">Now let [latex]g(x)=2x^3[\/latex], so [latex]g^{\\prime}(x)=6x^{2}[\/latex]. Substituting into the previous result, we obtain<\/p>\r\n\r\n<div id=\"fs-id1169739336080\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} h^{\\prime}(x) &amp; =\\dfrac{1}{\\sqrt{1-4x^6}} \\cdot 6x^{2} \\\\ &amp; =\\dfrac{6x^{2}}{\\sqrt{1-4x^6}} \\end{array}[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[\/hidden-answer]<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Applying the Chain Rule to the Inverse Sine Function.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/0dlA5QJZYsw?controls=0&amp;start=628&amp;end=723&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\/3.7DerivativesOfInverseFunctions628to723_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-id1169739303911\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169739111210\">Use the inverse function theorem to find the derivative of [latex]g(x)=\\tan^{-1} x[\/latex].<\/p>\r\n[reveal-answer q=\"462877\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"462877\"]\r\n<p id=\"fs-id1169736613675\">The inverse of [latex]g(x)[\/latex] is [latex]f(x)= \\tan x[\/latex]. Use <a class=\"autogenerated-content\" href=\"#fs-id1169738977168\">(Figure)<\/a> as a guide.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169736659262\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169736659262\"]\r\n<p id=\"fs-id1169736659262\">[latex]g^{\\prime}(x)=\\dfrac{1}{1+x^2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1169739282743\">The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. These formulas are provided in the following theorem.<\/p>\r\n\r\n<div id=\"fs-id1169739282748\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Derivatives of Inverse Trigonometric Functions<\/h3>\r\n\r\n<hr \/>\r\n\r\n<div id=\"fs-id1169739303820\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}\\frac{d}{dx}(\\sin^{-1} x)=\\large \\frac{1}{\\sqrt{1-x^2}} &amp; &amp; &amp; &amp; \\frac{d}{dx}(\\cos^{-1} x)=\\large \\frac{-1}{\\sqrt{1-x^2}} \\\\ \\frac{d}{dx}(\\tan^{-1} x)=\\large \\frac{1}{1+x^2} &amp; &amp; &amp; &amp; \\frac{d}{dx}(\\cot^{-1} x)=\\large \\frac{-1}{1+x^2} \\\\ \\frac{d}{dx}(\\sec^{-1} x)=\\large \\frac{1}{|x|\\sqrt{x^2-1}} &amp; &amp; &amp; &amp; \\frac{d}{dx}(\\csc^{-1} x)=\\large \\frac{-1}{|x|\\sqrt{x^2-1}} \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<div class=\"equation\"><\/div>\r\n<\/div>\r\n<div class=\"textbook exercises\">\r\n<h3>Example: Applying Differentiation Formulas to an Inverse Tangent Function<\/h3>\r\n<p id=\"fs-id1169736654830\">Find the derivative of [latex]f(x)=\\tan^{-1} (x^2)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169739270394\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739270394\"]\r\n<p id=\"fs-id1169739270394\">Let [latex]g(x)=x^2[\/latex], so [latex]g^{\\prime}(x)=2x[\/latex]. Substituting into <a class=\"autogenerated-content\" href=\"#fs-id1169739307948\">(Figure)<\/a>, we obtain<\/p>\r\n\r\n<div id=\"fs-id1169736661176\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{1}{1+(x^2)^2} \\cdot (2x)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736595973\">Simplifying, we have<\/p>\r\n\r\n<div id=\"fs-id1169736595976\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{2x}{1+x^4}[\/latex]<\/div>\r\n&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169739301501\" class=\"textbook exercises\">\r\n<h3>Example: Applying Differentiation Formulas to an Inverse Sine Function<\/h3>\r\n<p id=\"fs-id1169739027843\">Find the derivative of [latex]h(x)=x^2 \\sin^{-1} x[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1169736603526\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169736603526\"]\r\n<p id=\"fs-id1169736603526\">By applying the product rule, we have<\/p>\r\n\r\n<div id=\"fs-id1169736603529\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=2x \\sin^{-1} x+\\frac{1}{\\sqrt{1-x^2}} \\cdot x^2[\/latex].<\/div>\r\n&nbsp;\r\n<div class=\"equation unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Applying Differentiation Formulas to an Inverse Sine Function.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/0dlA5QJZYsw?controls=0&amp;start=786&amp;end=848&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\/3.7DerivativesOfInverseFuunctions786to848_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-id1169739188429\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169739325671\">Find the derivative of [latex]h(x)= \\cos^{-1} (3x-1)[\/latex]<\/p>\r\n[reveal-answer q=\"27761093\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"27761093\"]\r\n<p id=\"fs-id1169739274281\">Use <a class=\"autogenerated-content\" href=\"#fs-id1169736659297\">(Figure)<\/a>. with [latex]g(x)=3x-1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739304017\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739304017\"]\r\n<p id=\"fs-id1169739304017\">[latex]h^{\\prime}(x)=\\frac{-3}{\\sqrt{6x-9x^2}}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169736614197\" class=\"textbook exercises\">\r\n<h3>Example: Applying the Inverse Tangent Function<\/h3>\r\n<p id=\"fs-id1169739208500\">The position of a particle at time [latex]t[\/latex] is given by [latex]s(t)= \\tan^{-1}\\left(\\dfrac{1}{t}\\right)[\/latex] for [latex]t\\ge \\frac{1}{2}[\/latex]. Find the velocity of the particle at time [latex]t=1[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1169736656603\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169736656603\"]\r\n<p id=\"fs-id1169736656603\">Begin by differentiating [latex]s(t)[\/latex] in order to find [latex]v(t)[\/latex]. Thus,<\/p>\r\n\r\n<div id=\"fs-id1169739242485\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(t)=s^{\\prime}(t)=\\dfrac{1}{1+(\\frac{1}{t})^2} \\cdot \\dfrac{-1}{t^2}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739282656\">Simplifying, we have<\/p>\r\n\r\n<div id=\"fs-id1169736615257\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(t)=-\\dfrac{1}{t^2+1}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736594230\" style=\"text-align: center;\">Thus, [latex]v(1)=-\\dfrac{1}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169736605027\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169739273648\">Find the equation of the line tangent to the graph of [latex]f(x)= \\sin^{-1} x[\/latex] at [latex]x=0[\/latex].<\/p>\r\n[reveal-answer q=\"2340987\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"2340987\"]\r\n<p id=\"fs-id1169739299514\">[latex]f^{\\prime}(0)[\/latex] gives the slope of the tangent line.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739302752\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739302752\"]\r\n<p id=\"fs-id1169739302752\">[latex]y=x[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]206679[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Recognize the derivatives of the standard inverse trigonometric functions<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1169739301198\" class=\"bc-section section\">\n<p id=\"fs-id1169739029363\">We now turn our attention to finding derivatives of inverse trigonometric functions. These derivatives will prove invaluable in the study of integration later in this text. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are actually algebraic functions. Previously, derivatives of algebraic functions have proven to be algebraic functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Here, for the first time, we see that the derivative of a function need not be of the same type as the original function.<\/p>\n<div id=\"fs-id1169739208944\" class=\"textbook exercises\">\n<h3>Example: Derivative of the Inverse Sine Function<\/h3>\n<p id=\"fs-id1169739188589\">Use the inverse function theorem to find the derivative of [latex]g(x)=\\sin^{-1} x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739269789\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739269789\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739269789\">Since for [latex]x[\/latex] in the interval [latex][-\\frac{\\pi}{2},\\frac{\\pi}{2}], \\, f(x)= \\sin x[\/latex] is the inverse of [latex]g(x)= \\sin^{-1} x[\/latex], begin by finding [latex]f^{\\prime}(x)[\/latex]. Since<\/p>\n<div id=\"fs-id1169736611678\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)= \\cos x[\/latex] and [latex]f^{\\prime}(g(x))= \\cos (\\sin^{-1} x)=\\sqrt{1-x^2}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739347090\">we see that<\/p>\n<div id=\"fs-id1169739369239\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g^{\\prime}(x)=\\frac{d}{dx}(\\sin^{-1} x)=\\frac{1}{f^{\\prime}(g(x))}=\\frac{1}{\\sqrt{1-x^2}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<h4>Analysis<\/h4>\n<p id=\"fs-id1169739111057\">To see that [latex]\\cos (\\sin^{-1} x)=\\sqrt{1-x^2}[\/latex], consider the following argument. Set [latex]\\sin^{-1} x=\\theta[\/latex]. In this case, [latex]\\sin \\theta =x[\/latex] where [latex]-\\frac{\\pi}{2}\\le \\theta \\le \\frac{\\pi}{2}[\/latex]. We begin by considering the case where [latex]0<\\theta <\\frac{\\pi}{2}[\/latex]. Since [latex]\\theta[\/latex] is an acute angle, we may construct a right triangle having acute angle [latex]\\theta[\/latex], a hypotenuse of length 1, and the side opposite angle [latex]\\theta[\/latex] having length [latex]x[\/latex]. From the Pythagorean theorem, the side adjacent to angle [latex]\\theta[\/latex] has length [latex]\\sqrt{1-x^2}[\/latex]. This triangle is shown in Figure 2. Using the triangle, we see that [latex]\\cos (\\sin^{-1} x)= \\cos \\theta =\\sqrt{1-x^2}[\/latex].<\/p>\n<div style=\"width: 436px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205436\/CNX_Calc_Figure_03_07_002.jpg\" alt=\"A right triangle with angle \u03b8, opposite side x, hypotenuse 1, and adjacent side equal to the square root of the quantity (1 \u2013 x2).\" width=\"426\" height=\"205\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. Using a right triangle having acute angle [latex]\\theta[\/latex], a hypotenuse of length 1, and the side opposite angle [latex]\\theta[\/latex] having length [latex]x[\/latex], we can see that [latex]\\cos (\\sin^{-1} x)= \\cos \\theta =\\sqrt{1-x^2}[\/latex].<\/p>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1169739303300\" class=\"equation unnumbered\">In the case where [latex]-\\frac{\\pi}{2}<\\theta <0[\/latex], we make the observation that [latex]0<-\\theta<\\frac{\\pi}{2}[\/latex] and hence [latex]\\cos (\\sin^{-1} x)= \\cos \\theta = \\cos (\u2212\\theta )=\\sqrt{1-x^2}[\/latex].<\/div>\n<div><\/div>\n<p id=\"fs-id1169736611289\">Now if [latex]\\theta =\\frac{\\pi}{2}[\/latex] or [latex]\\theta =-\\frac{\\pi}{2}, \\, x=1[\/latex] or [latex]x=-1[\/latex], and since in either case [latex]\\cos \\theta =0[\/latex] and [latex]\\sqrt{1-x^2}=0[\/latex], we have<\/p>\n<div id=\"fs-id1169739340290\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\cos (\\sin^{-1} x)= \\cos \\theta =\\sqrt{1-x^2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739190632\">Consequently, in all cases, [latex]\\cos (\\sin^{-1} x)=\\sqrt{1-x^2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169736662939\" class=\"textbook exercises\">\n<h3>Example: Applying the Chain Rule to the Inverse Sine Function<\/h3>\n<p id=\"fs-id1169739190553\">Apply the chain rule to the formula derived in <em>Example: Applying the Inverse Function Theorem<\/em><br \/>\nto find the derivative of [latex]h(x)=\\sin^{-1} (g(x))[\/latex] and use this result to find the derivative of [latex]h(x)=\\sin^{-1}(2x^3)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739189899\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739189899\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739189899\">Applying the chain rule to [latex]h(x)=\\sin^{-1} (g(x))[\/latex], we have<\/p>\n<div id=\"fs-id1169736596017\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=\\dfrac{1}{\\sqrt{1-(g(x))^2}}g^{\\prime}(x)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739269950\">Now let [latex]g(x)=2x^3[\/latex], so [latex]g^{\\prime}(x)=6x^{2}[\/latex]. Substituting into the previous result, we obtain<\/p>\n<div id=\"fs-id1169739336080\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} h^{\\prime}(x) & =\\dfrac{1}{\\sqrt{1-4x^6}} \\cdot 6x^{2} \\\\ & =\\dfrac{6x^{2}}{\\sqrt{1-4x^6}} \\end{array}[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Applying the Chain Rule to the Inverse Sine Function.<\/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=628&amp;end=723&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\/3.7DerivativesOfInverseFunctions628to723_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-id1169739303911\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169739111210\">Use the inverse function theorem to find the derivative of [latex]g(x)=\\tan^{-1} x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q462877\">Hint<\/span><\/p>\n<div id=\"q462877\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736613675\">The inverse of [latex]g(x)[\/latex] is [latex]f(x)= \\tan x[\/latex]. Use <a class=\"autogenerated-content\" href=\"#fs-id1169738977168\">(Figure)<\/a> as a guide.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169736659262\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169736659262\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736659262\">[latex]g^{\\prime}(x)=\\dfrac{1}{1+x^2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169739282743\">The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. These formulas are provided in the following theorem.<\/p>\n<div id=\"fs-id1169739282748\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Derivatives of Inverse Trigonometric Functions<\/h3>\n<hr \/>\n<div id=\"fs-id1169739303820\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{array}{lllll}\\frac{d}{dx}(\\sin^{-1} x)=\\large \\frac{1}{\\sqrt{1-x^2}} & & & & \\frac{d}{dx}(\\cos^{-1} x)=\\large \\frac{-1}{\\sqrt{1-x^2}} \\\\ \\frac{d}{dx}(\\tan^{-1} x)=\\large \\frac{1}{1+x^2} & & & & \\frac{d}{dx}(\\cot^{-1} x)=\\large \\frac{-1}{1+x^2} \\\\ \\frac{d}{dx}(\\sec^{-1} x)=\\large \\frac{1}{|x|\\sqrt{x^2-1}} & & & & \\frac{d}{dx}(\\csc^{-1} x)=\\large \\frac{-1}{|x|\\sqrt{x^2-1}} \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div class=\"equation\"><\/div>\n<\/div>\n<div class=\"textbook exercises\">\n<h3>Example: Applying Differentiation Formulas to an Inverse Tangent Function<\/h3>\n<p id=\"fs-id1169736654830\">Find the derivative of [latex]f(x)=\\tan^{-1} (x^2)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739270394\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739270394\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739270394\">Let [latex]g(x)=x^2[\/latex], so [latex]g^{\\prime}(x)=2x[\/latex]. Substituting into <a class=\"autogenerated-content\" href=\"#fs-id1169739307948\">(Figure)<\/a>, we obtain<\/p>\n<div id=\"fs-id1169736661176\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{1}{1+(x^2)^2} \\cdot (2x)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736595973\">Simplifying, we have<\/p>\n<div id=\"fs-id1169736595976\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{2x}{1+x^4}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169739301501\" class=\"textbook exercises\">\n<h3>Example: Applying Differentiation Formulas to an Inverse Sine Function<\/h3>\n<p id=\"fs-id1169739027843\">Find the derivative of [latex]h(x)=x^2 \\sin^{-1} x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169736603526\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169736603526\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736603526\">By applying the product rule, we have<\/p>\n<div id=\"fs-id1169736603529\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]h^{\\prime}(x)=2x \\sin^{-1} x+\\frac{1}{\\sqrt{1-x^2}} \\cdot x^2[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<div class=\"equation unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Applying Differentiation Formulas to an Inverse Sine Function.<\/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=786&amp;end=848&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\/3.7DerivativesOfInverseFuunctions786to848_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-id1169739188429\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169739325671\">Find the derivative of [latex]h(x)= \\cos^{-1} (3x-1)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q27761093\">Hint<\/span><\/p>\n<div id=\"q27761093\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739274281\">Use <a class=\"autogenerated-content\" href=\"#fs-id1169736659297\">(Figure)<\/a>. with [latex]g(x)=3x-1[\/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-id1169739304017\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739304017\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739304017\">[latex]h^{\\prime}(x)=\\frac{-3}{\\sqrt{6x-9x^2}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169736614197\" class=\"textbook exercises\">\n<h3>Example: Applying the Inverse Tangent Function<\/h3>\n<p id=\"fs-id1169739208500\">The position of a particle at time [latex]t[\/latex] is given by [latex]s(t)= \\tan^{-1}\\left(\\dfrac{1}{t}\\right)[\/latex] for [latex]t\\ge \\frac{1}{2}[\/latex]. Find the velocity of the particle at time [latex]t=1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169736656603\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169736656603\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169736656603\">Begin by differentiating [latex]s(t)[\/latex] in order to find [latex]v(t)[\/latex]. Thus,<\/p>\n<div id=\"fs-id1169739242485\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(t)=s^{\\prime}(t)=\\dfrac{1}{1+(\\frac{1}{t})^2} \\cdot \\dfrac{-1}{t^2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739282656\">Simplifying, we have<\/p>\n<div id=\"fs-id1169736615257\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]v(t)=-\\dfrac{1}{t^2+1}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736594230\" style=\"text-align: center;\">Thus, [latex]v(1)=-\\dfrac{1}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169736605027\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169739273648\">Find the equation of the line tangent to the graph of [latex]f(x)= \\sin^{-1} x[\/latex] at [latex]x=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q2340987\">Hint<\/span><\/p>\n<div id=\"q2340987\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739299514\">[latex]f^{\\prime}(0)[\/latex] gives the slope of the tangent line.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739302752\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739302752\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739302752\">[latex]y=x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm206679\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=206679&theme=oea&iframe_resize_id=ohm206679&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-378\">\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":31,"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-378","chapter","type-chapter","status-publish","hentry"],"part":35,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/378","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":22,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/378\/revisions"}],"predecessor-version":[{"id":4817,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/378\/revisions\/4817"}],"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\/378\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=378"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=378"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=378"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=378"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}