Differentiate the following functions with respect to $x$ :
$e^{\tan 3 x}$
Let $y=e^{\tan 3 x}$
On differentiating $y$ with respect to $x$, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(e^{\tan 3 x}\right)$
We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\mathrm{x}}\right)=\mathrm{e}^{\mathrm{x}}$
$\Rightarrow \frac{d y}{d x}=e^{\tan 3 x} \frac{d}{d x}(\tan 3 x)$ [using chain rule]
We have $\frac{d}{d x}(\tan x)=\sec ^{2} x$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\tan 3 \mathrm{x}} \sec ^{2} 3 \mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}(3 \mathrm{x})$ [using chain rule]
$\Rightarrow \frac{d y}{d x}=3 e^{\tan 3 x} \sec ^{2} 3 x \frac{d}{d x}(x)$
However, $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1$
$\Rightarrow \frac{d y}{d x}=3 e^{\tan 3 x} \sec ^{2} 3 x \times 1$
$\therefore \frac{d y}{d x}=3 e^{\tan 3 x} \sec ^{2} 3 x$
Thus, $\frac{d}{d x}\left(e^{\tan 3 x}\right)=3 e^{\tan 3 x} \sec ^{2} 3 x$