Find the derivative of f given by f(x) = sin^ | vedclass

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Question

(A) Let $y = \sin^{-1} x$. Then,$x = \sin y$. Differentiating both sides with respect to $x$,we get: $1 = \cos y \frac{dy}{dx}$ This implies that $\fr

Vedclass · Accepted Answer

$\frac{1}{\sqrt{1-x^2}}$

Vedclass · Answer

(A) Let $y = \sin^{-1} x$. Then,$x = \sin y$. Differentiating both sides with respect to $x$,we get: $1 = \cos y \frac{dy}{dx}$ This implies that $\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\cos(\sin^{-1} x)}$. Observe that this is defined only for $\cos y 
eq 0$,i.e.,$y 
eq \pm \frac{\pi}{2}$,which means $x 
eq \pm 1$,or $x \in (-1, 1)$. Since $\cos^2 y = 1 - \sin^2 y = 1 - x^2$ and $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$,$\cos y$ must be positive. Therefore,$\cos y = \sqrt{1 - x^2}$. Thus,for $x \in (-1, 1)$,$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$.

Find the derivative of $f$ given by $f(x) = \sin^{-1} x$ assuming it exists.

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