Differentiate the following functions with respect to x :

Let $y=\tan ^{-1}\left(e^{x}\right)$

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

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

We know $\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}$

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

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

However, $\frac{d}{d x}\left(e^{x}\right)=e^{x}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{1+e^{2 x}} \times e^{x}$

$\therefore \frac{d y}{d x}=\frac{e^{x}}{1+e^{2 x}}$

Thus, $\frac{d}{d x}\left(\tan ^{-1} e^{x}\right)=\frac{e^{x}}{1+e^{2 x}}$