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Question:

If $y=\log \sqrt{\frac{1+\tan x}{1-\tan x}}$, prove that $\frac{d y}{d x}=\sec 2 x$

Solution:

Given $y=\log \sqrt{\frac{1+\tan x}{1-\tan x}}$

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

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

We know $\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})=\frac{1}{\mathrm{x}}$

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

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

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

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

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

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

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

Recall that $\left(\frac{\mathrm{u}}{\mathrm{v}}\right)^{\prime}=\frac{\mathrm{vu}^{\prime}-\mathrm{uv}^{\prime}}{\mathrm{v}^{2}}$ (quotient rule)

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$

$=\frac{1}{2}\left(\frac{1-\tan x}{1+\tan x}\right)\left[\frac{(1-\tan x)\left(\frac{d}{d x}(1)+\frac{d}{d x}(\tan x)\right)-(1+\tan x)\left(\frac{d}{d x}(1)-\frac{d}{d x}(\tan x)\right)}{(1-\tan x)^{2}}\right]$

We know $\frac{d}{d x}(\tan x)=\sec ^{2} x$ and derivative of a constant is 0 .

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left(\frac{1-\tan x}{1+\tan x}\right)\left[\frac{(1-\tan x)\left(0+\sec ^{2} x\right)-(1+\tan x)\left(0-\sec ^{2} x\right)}{(1-\tan x)^{2}}\right]$

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

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

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

$\Rightarrow \frac{d y}{d x}=\frac{\sec ^{2} x}{(1+\tan x)(1-\tan x)}$

$\Rightarrow \frac{d y}{d x}=\frac{\sec ^{2} x}{1-\tan ^{2} x}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1+\tan ^{2} \mathrm{x}}{1-\tan ^{2} \mathrm{x}}\left(\because \sec ^{2} \theta-\tan ^{2} \theta=1\right)$

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

$\Rightarrow \frac{d y}{d x}=\frac{\cos ^{2} x+\sin ^{2} x}{\cos ^{2} x-\sin ^{2} x}$

But, $\cos ^{2} \theta+\sin ^{2} \theta=1$ and $\cos ^{2} \theta-\sin ^{2} \theta=\cos (2 \theta)$.

$\Rightarrow \frac{d y}{d x}=\frac{1}{\cos 2 x}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\sec 2 \mathrm{x}$

Thus, $\frac{\mathrm{dy}}{\mathrm{dx}}=\sec 2 \mathrm{x}$

 

 

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