Question

Differentiate

${\cos }^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right),-1<x<1$

Answer 1

$$\begin{gathered} \mathrm{Let}y={\cos }^{-1}\left\{\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right\} \\ \mathrm{Put},x=\cos \theta \\ y={\cos }^{-1}\left\{\frac{\cos \theta +\sqrt{1-{\cos }^2\theta }}{\sqrt{2}}\right\} \\ y={\cos }^{-1}\left\{\frac{\cos \theta +\sin \theta }{\sqrt{2}}\right\} \\ y={\cos }^{-1}\left\{\cos \theta \left(\frac{1}{\sqrt{2}}\right)+\sin \theta \left(\frac{1}{\sqrt{2}}\right)\right\} \\ y={\cos }^{-1}\left\{\cos \theta \cos \frac{\mathrm{\pi }}{4}+\sin \theta \sin \frac{\mathrm{\pi }}{4}\right\} \\ y={\cos }^{-1}\left\{\cos \left(\theta -\frac{\mathrm{\pi }}{4}\right)\right\}...\left(\mathrm{i}\right) \\ \mathrm{Here},-1<x<1 \\ \Rightarrow -1<\cos \theta <1 \\ \Rightarrow \frac{3\mathrm{\pi }}{4}<\theta <\frac{5\mathrm{\pi }}{4} \\ \Rightarrow \left(\frac{3\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}\right)<\left(\theta -\frac{\mathrm{\pi }}{4}\right)<\frac{5\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4} \\ \Rightarrow \left(\frac{\mathrm{\pi }}{2}\right)<\left(\theta -\frac{\mathrm{\pi }}{4}\right)<\mathrm{\pi } \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ y=\left(\theta -\frac{\mathrm{\pi }}{4}\right)\left[\mathrm{Since},{\cos }^{-1}\left(\cos \theta \right)=\theta ,\mathrm{if}\theta \in \left[0,\mathrm{\pi }\right]\right] \\ y={\cos }^{-1}x-\frac{\mathrm{\pi }}{4}\left[\mathrm{Since},x=\sin \theta \right] \\ \mathrm{Differentiating}\mathrm{it}\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}+0 \\ \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}} \end{gathered}$$