Differentiate the following functions with respect to x :

Let $y=e^{\sin ^{-1} 2 x}$

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

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

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\mathrm{x}}\right)=\mathrm{e}^{\mathrm{x}}$

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

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

$\Rightarrow \frac{d y}{d x}=e^{\sin ^{-1} 2 x} \frac{1}{\sqrt{1-(2 x)^{2}}} \frac{d}{d x}(2 x)$ [using chain rule]

$\Rightarrow \frac{d y}{d x}=\frac{e^{\sin ^{-1} 2 x}}{\sqrt{1-4 x^{2}}} \times 2 \frac{d}{d x}(x)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{e}^{\sin ^{-1} 2 \mathrm{x}}}{\sqrt{1-4 \mathrm{x}^{2}}} \times \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$

However, $\frac{d}{d x}(x)=1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{e}^{\sin ^{-1} 2 \mathrm{x}}}{\sqrt{1-4 \mathrm{x}^{2}}} \times 1$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{e}^{\sin ^{-1} 2 \mathrm{x}}}{\sqrt{1-4 \mathrm{x}^{2}}}$

Thus, $\frac{d}{d x}\left(e^{\sin ^{-1} 2 x}\right)=\frac{2 e^{\sin ^{-1} 2 x}}{\sqrt{1-4 x^{2}}}$