Differentiate the following functions with respect to x :

Solution:

$y=\cos ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\}$

Let $x=\sin \theta$

Now

$y=\cos ^{-1}\left\{\frac{\sin \theta+\sqrt{1-\sin ^{2} \theta}}{\sqrt{2}}\right\}$

Using $\sin ^{2} \theta+\cos ^{2} \theta=1$

$y=\cos ^{-1}\left\{\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right\}$

Now

$y=\cos ^{-1}\left\{\sin \theta \frac{1}{\sqrt{2}}+\cos \theta \frac{1}{\sqrt{2}}\right\}$

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

Using $\cos (A-B)=\cos A \cos B+\sin A \sin B$

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

Considering the limits,

$-1

$-1<\sin \theta<1$

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

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

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

Now,

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

$y=-\left(\theta-\frac{\pi}{4}\right)$

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

Differentiating w.r.t $x$, we get

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

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