#### provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 53

Answer: $-\sec x$

Hint: you must know the rule of solving derivative of logarithm and trigonometric functions

Given: $\log \left\{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}$

Solution:

Let  $y=\log \left\{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}$

Differentiate with respect to x

$\frac{d y}{d x}=\frac{1}{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)} \cdot \frac{d}{d x} \cot \left(\frac{x}{2}+\frac{\pi}{4}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)} \cdot-\operatorname{cosec}^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right) \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{4}+\frac{x}{2}\right)$

\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\operatorname{cosec}^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}{\cot \left(\frac{\pi}{4}+\frac{x}{2}\right)} \times \frac{1}{2} \\\\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\operatorname{cosec}^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}{2 \cot \left(\frac{\pi}{4}+\frac{x}{2}\right)} \end{aligned}

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{\sin ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)} \times \frac{\sin \left(\frac{\pi}{4}+\frac{x}{2}\right)}{2 \cos \left(\frac{\pi}{4}+\frac{x}{2}\right)}$

$\frac{d y}{d x}=\frac{-1}{2 \cos \left(\frac{\pi}{4}+\frac{x}{2}\right) \sin \left(\frac{\pi}{4}+\frac{x}{2}\right)}$                    $[\therefore 2 \sin x \cos x=\sin 2 x]$

$\frac{d y}{d x}=\frac{-1}{\sin \left(\frac{\pi}{2}+x\right)}$

\begin{aligned} &\frac{d y}{d x}=-\frac{1}{\cos x} \\\\ &\frac{d y}{d x}=-\sec x \end{aligned}