{"id":258,"date":"2023-09-20T22:48:37","date_gmt":"2023-09-20T22:48:37","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/derivatives-of-inverse-trigonometric-functions\/"},"modified":"2024-08-05T12:53:40","modified_gmt":"2024-08-05T12:53:40","slug":"derivatives-of-inverse-functions-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/derivatives-of-inverse-functions-learn-it-2\/","title":{"raw":"Derivatives of Inverse Functions: Learn It 2","rendered":"Derivatives of Inverse Functions: Learn It 2"},"content":{"raw":"

Derivatives of Inverse Trigonometric Functions<\/h2>\r\n

We now shift our focus to the derivatives of inverse trigonometric functions, which play a crucial role in the study of integration later in this course. Intriguingly, unlike their trigonometric counterparts, the derivatives of inverse trigonometric functions are algebraic. This is a notable deviation from previous patterns observed, where derivatives of algebraic functions typically remained algebraic, and derivatives of trigonometric functions were also trigonometric. This departure underscores an important mathematical insight: the derivative of a function does not necessarily share the same type as the original function.<\/p>\r\n

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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\"]<\/p>\r\n

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].<\/p>\r\n

Since:<\/p>\r\n

[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

we see that:<\/p>\r\n

[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

Analysis<\/strong><\/p>\r\n

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>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"426\"]\"A 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\r\n

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>\r\n

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

[latex]\\cos (\\sin^{-1} x)= \\cos \\theta =\\sqrt{1-x^2}[\/latex]<\/div>\r\n

Consequently, in all cases, [latex]\\cos (\\sin^{-1} x)=\\sqrt{1-x^2}[\/latex].<\/p>\r\n

[\/hidden-answer]<\/p>\r\n<\/section>\r\n

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Apply the chain rule to 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\"]<\/p>\r\n

Applying the chain rule to [latex]h(x)=\\sin^{-1} (g(x))[\/latex], we have<\/p>\r\n

[latex]h^{\\prime}(x)=\\dfrac{1}{\\sqrt{1-(g(x))^2}}g^{\\prime}(x)[\/latex].<\/div>\r\n

 <\/p>\r\n

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

[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>\r\n

Watch the following video to see the worked solution to this example.<\/p>\r\n