#### need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 62

Hint:: you must know the rules of solving derivation of logarithm functions.

Given:$y=\log \left(\sqrt{\frac{1+\tan x}{1-\tan x}}\right)$

Prove   $\frac{d y}{d x}=\sec 2 x$

Solution:

Let  $y=\log \left(\sqrt{\frac{1+\tan x}{1-\tan x}}\right)$

$\mathrm{y}=\log \left(\frac{1+\tan x}{1-\tan x}\right)^{\frac{1}{2}}$                    $\left[\therefore \log a^{x}=x \log a\right]$

$\mathrm{y}=\frac{1}{2}[\log (1+\tan \mathrm{x})-\log (1-\tan \mathrm{x})]$            $\left[\therefore \log \frac{m}{n}=\log m-\log n\right]$

Differentiate with respect to x,

\begin{aligned} &\frac{d y}{d x}=\frac{1}{2}\left\{\frac{d}{d x}\left[\log (1+\tan x)-\frac{d}{d x} \log (1-\tan x)\right]\right\} \\\\ &\frac{d y}{d x}=\frac{1}{2}\left\{\frac{1}{1+\tan x} \frac{d}{d x}(1+\tan x)-\frac{1}{1-\tan x} \frac{d}{d x}(1-\tan x)\right\} \end{aligned}

\begin{aligned} &\frac{d y}{d x}=\frac{1}{2}\left\{\frac{1}{1+\tan x}\left(\sec ^{2} x\right)-\frac{1}{1-\tan x}\left(-\sec ^{2} x\right)\right\} \\\\ &\frac{d y}{d x}=\frac{1}{2}\left\{\frac{\left(\sec ^{2} x\right)}{1+\tan x}+\frac{\sec ^{2} x}{1-\tan x}\right\} \end{aligned}

\begin{aligned} &\frac{d y}{d x}=\frac{\sec ^{2} x}{2}\left[\frac{1-\tan x+1+\tan x}{1-\tan ^{2} x}\right] \\\\ &\frac{d y}{d x}=\frac{1}{2} \sec ^{2} x\left[\frac{2}{1-\tan ^{2} x}\right] \end{aligned}

$\frac{d y}{d x}=\frac{\sec ^{2} x}{1-\tan ^{2} x}$                    $\left[\therefore \sec ^{2} x=1+\tan ^{2} x\right]$

\begin{aligned} &\frac{d y}{d x}=\frac{1+\tan ^{2} x}{1-\tan ^{2} x} \\\\ &\frac{d y}{d x}=\frac{1}{\left(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right)} \end{aligned}                    $\left[\therefore \text { Tigonometric Property } \frac{1-\tan ^{2} x}{1+\tan ^{2} \mathrm{x}}=\cos 2 x\right]$

\begin{aligned} &\frac{d y}{d x}=\frac{1}{\cos 2 \mathrm{x}} \\\\ &\frac{d y}{d x}=\sec 2 x \end{aligned}

∴ Proved