Differentiate the following functions with respect to x :

Solution:

$y=\sin ^{-1}\left\{2 x^{2}-1\right\}$

let $\mathrm{x}=\cos \theta$

Now

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

Using $2 \cos ^{2} \theta-1=\cos 2 \theta$

$y=\sin ^{-1}(\cos 2 \theta)$

$y=\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-2 \theta\right)\right\}$

Considering the limits,

$0

$0<\cos \theta<1$

$0<\theta<\frac{\pi}{2}$

$0<2 \theta<\pi$

$0>-2 \theta>-\pi$

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

Now,

$y=\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-2 \theta\right)\right\}$

$y=\frac{\pi}{2}-2 \theta$

$y=\frac{\pi}{2}-2 \cos ^{-1} x$

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

$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{2}-2 \cos ^{-1} x\right)$

$\frac{d y}{d x}=0-2\left(-\frac{1}{\sqrt{1-x^{2}}}\right)$

$\frac{d y}{d x}=\frac{2}{\sqrt{1-x^{2}}}$