# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

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

$\Rightarrow \frac{1}{\sqrt{2}}<\cos \theta<1$

$\Rightarrow 0<\theta<\frac{\pi}{4}$

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

$\Rightarrow 0>-2 \theta>-\frac{\pi}{2}$

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

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

Therefore,

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

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

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

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{2}-2 \cos ^{-1} \mathrm{x}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=0-2\left(\frac{-1}{\sqrt{1-\mathrm{x}^{2}}}\right)$

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