# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

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

Solution:

Let $\mathrm{y}=\mathrm{e}^{\sqrt{\cot \mathrm{x}}}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\sqrt{\cot \mathrm{x}}}\right)$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\sqrt{\cot x}} \frac{\mathrm{d}}{\mathrm{dx}}(\sqrt{\cot x})$ [using chain rule]

$\Rightarrow \frac{d y}{d x}=e^{\sqrt{\cot x}} \frac{d}{d x}\left[(\cot x)^{\frac{1}{2}}\right]$

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

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

$\Rightarrow \frac{d y}{d x}=\frac{1}{2} e^{\sqrt{\cot x}}(\cot x)^{-\frac{1}{2}} \frac{d}{d x}(\cot x)$

However, $\frac{d}{d x}(\cot x)=-\operatorname{cosec}^{2} x$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{2} \mathrm{e}^{\sqrt{\cot \mathrm{x}}}(\cot \mathrm{x})^{-\frac{1}{2}} \operatorname{cosec}^{2} \mathrm{x}$

$\Rightarrow \frac{d y}{d x}=-\frac{e^{\sqrt{\cot x}} \operatorname{cosec}^{2} x}{2(\cot x)^{\frac{1}{2}}}$

$\therefore \frac{d y}{d x}=-\frac{e^{\sqrt{\cot x}} \operatorname{cosec}^{2} x}{2 \sqrt{\cot x}}$

Thus, $\frac{d}{d x}\left(e^{\sqrt{\cot x}}\right)=-\frac{e^{\sqrt{\cot x}} \operatorname{cosec}^{2} x}{2 \sqrt{\cot x}}$