Differentiate the following functions with respect to x :

Let $y=\tan \left(e^{\sin x}\right)$

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\tan \left(\mathrm{e}^{\sin \mathrm{x}}\right)\right]$

We know $\frac{d}{d x}(\tan x)=\sec ^{2} x$

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

We have $\frac{d}{d x}\left(e^{x}\right)=e^{x}$

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

However, $\frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})=\cos \mathrm{x}$

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

$\therefore \frac{d y}{d x}=e^{\sin x} \cos x \sec ^{2}\left(e^{\sin x}\right)$

Thus, $\frac{d}{d x}\left[\tan \left(e^{\sin x}\right)\right]=e^{\sin x} \cos x \sec ^{2}\left(e^{\sin x}\right)$