Question

Differentiate

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

Answer 1

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