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I can find this using the fact that $\sin(\sin^{-1}(x)) = x$, for all $x\in[-1,1].$

Now, differentiate.

$$\frac{d}{d\sin^{-1}(x)}\sin(\sin^{-1}(x))\cdot \frac{d}{dx} \sin^{-1}(x)= \frac{d}{dx} x= 1$$

$$\cos(\sin^{-1}(x))\cdot \frac{d}{dx} \sin^{-1}(x) = 1$$

$$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\cos(\sin^{-1}(x))}$$

$$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-\sin^2(\sin^{-1}(x))}}$$

$$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$$

However, what if I wanted to differentiate this like $\ \sin^{-1}(\sin(x))$ without knowing the fact that $\ \frac{d}{dx}\sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$ ? Is there a solution for it? I keep getting stuck at a certain step when I try this...

Kenta S
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user13123
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3 Answers3

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If you're asking how to differentiate$$y=\arcsin(\sin x)$$without using the fact that$$\frac {d}{dx}\,\arcsin x=\frac 1{\sqrt{1-x^2}}$$Then the easiest way is to use implicit differentiation. Setting your expression equal to $y$, we have that$$\sin y=\sin x$$So$$y_x\cdot\cos y=\cos x\quad\implies\quad y_x=\frac {\cos x}{\cos y}$$

Crescendo
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  • How would this work with $sin^{-1}x$? Do I need to set $sin^{-1}x = sin^{-1}y$? – user13123 Nov 23 '17 at 05:02
  • @user13123 I'm sorry, but I'm slightly confused what you're asking. If you're asked to differentiate a trigonometric inverse function, the first thing that comes to mind should be implicit differentiation. – Crescendo Nov 23 '17 at 05:23
  • Ok... but how would you implicitly differentiate $sin^{-1}x(sinx)$? – user13123 Nov 23 '17 at 05:39
  • @user13123 Well if we let $y=\arcsin(x\sin x)$ then$$\sin y=x\sin x$$ – Crescendo Nov 23 '17 at 17:45
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let $ y = \arcsin(\sin x) $. All you have to do then is consider $$ \sin y = \sin(\arcsin(\sin x))$$ and differentiate implicitly.

Kenta S
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Аlon Djurinsky
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You could do it like this: first, define $f$ by $f(x) = \sin^{-1}x,$ where $-1 \leq x \leq 1$ and $-\frac\pi2 \leq \sin^{-1}x \leq \frac\pi2.$ (Recall that $\sin^{-1}$ is not a real function for $|x| > 1$ and that $\sin$ is not a one-to-one function, so we have to choose an appropriate domain and range of $\sin^{-1}$; these are the usual choices.) Then for $-\frac\pi2 \leq \theta \leq \frac\pi2,$ $$ f(\sin \theta) = \theta. \tag1 $$ Take the derivative of each side of $(1)$ with respect to $\theta$: $$ \frac{d}{d\theta}\left(f(\sin\theta)\right) = 1. \tag2 $$ Evaluate the left-hand side of $(2)$ using the chain rule: $$ \frac{d}{d\theta}\left(f(\sin\theta)\right) = f'(\sin\theta)\frac{d}{d\theta}\left(\sin\theta\right) = f'(\sin\theta) \cos\theta. \tag3 $$ Combine $(2)$ and $(3)$: $$ f'(\sin\theta) \cos\theta = 1. \tag4 $$ Divide by $\cos\theta$ on both sides of $(4)$: $$ f'(\sin\theta) = \frac{1}{\cos\theta}. \tag5 $$ Let $x = \sin\theta.$ Then $1 - x^2 = \cos^2\theta,$ and since $-\frac\pi2 \leq \theta \leq \frac\pi2,$ it follows that $\cos\theta\geq 0,$ so we have $\cos\theta = \sqrt{1 - x^2}.$ Making these substitutions in $(5)$, $$ f'(x) = \frac{1}{\sqrt{1 - x^2}}. $$ But $f'(x)=\frac{d}{dx}\sin^{-1}x,$ so $$ \frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}. $$

David K
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