# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$e^{\sin \sqrt{x}}$

Solution:

Let $y=e^{\sin \sqrt{x}}$

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

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

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

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

We have $\frac{d}{d x}(\sin x)=\cos x$

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

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\sin \sqrt{\mathrm{x}}} \cos \sqrt{\mathrm{x}}\left[\frac{1}{2} \mathrm{x}\left(\frac{1}{2}-1\right)\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2} e^{\sin \sqrt{x}} \cos \sqrt{x} x^{-\frac{1}{2}}$

$\therefore \frac{d y}{d x}=\frac{1}{2 \sqrt{x}} e^{\sin \sqrt{x}} \cos \sqrt{x}$

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