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.
Use the inverse function theorem to find the derivative of [latex]g(x)=\sin^{-1} x[/latex].
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:
[latex]f^{\prime}(x)= \cos x[/latex] and [latex]f^{\prime}(g(x))= \cos (\sin^{-1} x)=\sqrt{1-x^2}[/latex],
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].
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].
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 (−\theta )=\sqrt{1-x^2}[/latex].
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
Consequently, in all cases, [latex]\cos (\sin^{-1} x)=\sqrt{1-x^2}[/latex].
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].
Applying the chain rule to [latex]h(x)=\sin^{-1} (g(x))[/latex], we have
Watch the following video to see the worked solution to this example.
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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.
Find the derivative of [latex]f(x)=\tan^{-1} (x^2)[/latex]
Let [latex]g(x)=x^2[/latex], so [latex]g^{\prime}(x)=2x[/latex]. Substituting into [latex]\frac{d}{dx}(\tan^{-1} x)=\large \frac{1}{1+x^2}[/latex], we obtain
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].
Begin by differentiating [latex]s(t)[/latex] in order to find [latex]v(t)[/latex]. Thus,
