# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$\sin ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\},-1 Solution:$y=\sin ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\}$Let$x=\sin \theta$Now$y=\sin ^{-1}\left\{\frac{\sin \theta+\sqrt{1-\sin ^{2} \theta}}{\sqrt{2}}\right\}$Using$\sin ^{2} \theta+\cos ^{2} \theta=1y=\sin ^{-1}\left\{\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right\}$Now$y=\sin ^{-1}\left\{\sin \theta \frac{1}{\sqrt{2}}+\cos \theta \frac{1}{\sqrt{2}}\right\}y=\sin ^{-1}\left\{\sin \theta \cos \left(\frac{\pi}{4}\right)+\cos \theta \sin \left(\frac{\pi}{4}\right)\right\}$Using$\sin (A+B)=\sin A \cos B+\cos A \sin By=\sin ^{-1}\left\{\sin \left(\theta+\frac{\pi}{4}\right)\right\}$Considering the limits,$-1

$-1<\sin \theta<1$

$-\frac{\pi}{2}<\theta<\frac{\pi}{2}$

$-\frac{\pi}{2}+\frac{\pi}{4}<\theta+\frac{\pi}{4}<\frac{\pi}{2}+\frac{\pi}{4}$

$-\frac{\pi}{4}<\theta+\frac{\pi}{4}<\frac{3 \pi}{4}$

Now,

$y=\sin ^{-1}\left\{\sin \left(\theta+\frac{\pi}{4}\right)\right\}$

$y=\theta+\frac{\pi}{4}$

$y=\sin ^{-1} x+\frac{\pi}{4}$

Differentiating w.r.t $\mathrm{x}$, we get

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}+\frac{\pi}{4}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$