Question

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

Answer 1

Now,

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

or, $y=\frac{1}{2}\log \frac{(1+\tan x)}{(1-\tan x)}$

or, $y=\frac{1}{2}\left(\log \right(1+\tan x)-\log (1-\tan x\left)\right)$

Now differentiating both sides with respect to $x$ we get,

$\frac{dy}{dx}=\frac{1}{2}(\frac{{\sec }^{2}x}{1+\tan x}-\frac{-{\sec }^{2}x}{1-\tan x})$

or, $\frac{dy}{dx}=\frac{1}{2}(\frac{{\sec }^{2}x}{1+\tan x}+\frac{{\sec }^{2}x}{1-\tan x})$

or, $\frac{dy}{dx}=\frac{1}{2}\left(\frac{2{\sec }^{2}x}{1-{\tan }^{2}x}\right)$

or, $\frac{dy}{dx}=\left(\frac{1+{\tan }^{2}x}{1-{\tan }^{2}x}\right)$

or, $\frac{dy}{dx}=\frac{1}{\cos 2x}$

or, $\frac{dy}{dx}=\sec 2x.$