Differentiate the following functions with respect to x :

Let $y=\left(\sin ^{-1} x^{4}\right)^{4}$

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

$\frac{d y}{d x}=\frac{d}{d x}\left[\left(\sin ^{-1} x^{4}\right)^{4}\right]$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=4\left(\sin ^{-1} \mathrm{x}^{4}\right)^{4-1} \frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}^{4}\right)$ [using chain rule]

$\Rightarrow \frac{d y}{d x}=4\left(\sin ^{-1} x^{4}\right)^{3} \frac{d}{d x}\left(\sin ^{-1} x^{4}\right)$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=4\left(\sin ^{-1} \mathrm{x}^{4}\right)^{3} \frac{1}{\sqrt{1-\left(\mathrm{x}^{4}\right)^{2}}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{4}\right)$ [using chain rule]

$\Rightarrow \frac{d y}{d x}=\frac{4\left(\sin ^{-1} x^{4}\right)^{3}}{\sqrt{1-x^{8}}} \frac{d}{d x}\left(x^{4}\right)$

We have $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{4}\right)=4 \mathrm{x}^{3}$

$\Rightarrow \frac{d y}{d x}=\frac{4\left(\sin ^{-1} x^{4}\right)^{3}}{\sqrt{1-x^{8}}} \times 4 x^{3}$

$\therefore \frac{d y}{d x}=\frac{16 x^{3}\left(\sin ^{-1} x^{4}\right)^{3}}{\sqrt{1-x^{8}}}$

Thus, $\frac{d}{d x}\left[\left(\sin ^{-1} x^{4}\right)^{4}\right]=\frac{16 x^{3}\left(\sin ^{-1} x^{4}\right)^{3}}{\sqrt{1-x^{8}}}$