The function y = sin^{-1}(cos x) is not diffe | vedclass

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(D) The function $y = \sin^{-1}(\cos x)$ is defined when $-1 \le \cos x \le 1$. However,the derivative of $\sin^{-1} u$ is $\frac{1}{\sqrt{1-u^2}}$,wh

Vedclass · Accepted Answer

All options are correct

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(D) The function $y = \sin^{-1}(\cos x)$ is defined when $-1 \le \cos x \le 1$. However,the derivative of $\sin^{-1} u$ is $\frac{1}{\sqrt{1-u^2}}$,which is undefined when $u = \pm 1$.Thus,the function $y = \sin^{-1}(\cos x)$ is not differentiable where $\cos x = 1$ or $\cos x = -1$.For $\cos x = -1$,$x = (2n+1)\pi$ for any integer $n$. For $n=0$,$x = \pi$.For $\cos x = 1$,$x = 2n\pi$ for any integer $n$. For $n=1$,$x = 2\pi$,and for $n=-1$,$x = -2\pi$.Since the function is not differentiable at $x = \pi$,$x = 2\pi$,and $x = -2\pi$,all the given options are correct.

The function $y = \sin^{-1}(\cos x)$ is not differentiable at . . . . . .

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