# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$\sin \left(2 \sin ^{-1} x\right)$

Solution:

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

On differentiating $y$ with respect to $x$, we get

$\frac{d y}{d x}=\frac{d}{d x}\left[\sin \left(2 \sin ^{-1} x\right)\right]$

We know $\frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})=\cos \mathrm{x}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\cos \left(2 \sin ^{-1} \mathrm{x}\right) \frac{\mathrm{d}}{\mathrm{dx}}\left(2 \sin ^{-1} \mathrm{x}\right)$ [using chain rule]

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

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

We have $\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}\right)=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$

$\Rightarrow \frac{d y}{d x}=2 \cos \left(2 \sin ^{-1} x\right) \times \frac{1}{\sqrt{1-x^{2}}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \cos \left(2 \sin ^{-1} \mathrm{x}\right)}{\sqrt{1-\mathrm{x}^{2}}}$

Thus, $\frac{d}{d x}\left[\sin \left(2 \sin ^{-1} x\right)\right]=\frac{2 \cos \left(2 \sin ^{-1} x\right)}{\sqrt{1-x^{2}}}$