Differentiate the following functions with respect to x :

Let $y=e^{\tan ^{-1} \sqrt{x}}$

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

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

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

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

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{e}^{\tan ^{-1} \sqrt{\mathrm{x}}}}{1+\mathrm{x}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\frac{1}{2}}\right)$

However, $\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$

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

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

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

$\therefore \frac{d y}{d x}=\frac{e^{\tan ^{-1} \sqrt{x}}}{2 \sqrt{x}(1+x)}$

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