Question

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

Answer 1

​$$\begin{gathered} \mathrm{Let}y=\log \sqrt{\frac{1+\tan x}{1-\tan x}} \\ \Rightarrow y=\log {\left\{\frac{1+\tan x}{1-\tan x}\right\}}^{\frac{1}{2}} \\ \Rightarrow y=\frac{1}{2}\log \left\{\frac{1+\tan x}{1-\tan x}\right\} \\ \Rightarrow y=\frac{1}{2}\left\{\log \left(1+\tan x\right)-\log \left(1-\tan x\right)\right\} \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{1}{2}\left\{\frac{d}{dx}\left\{\log \left(1+\tan x\right)\right\}-\frac{d}{dx}\left\{\log \left(1-\tan x\right)\right\}\right\} \\ =\frac{1}{2}\left\{\frac{1}{1+\tan x}\times \frac{d}{dx}\left(1+\tan x\right)-\frac{1}{1-\tan x}\times \frac{d}{dx}\left(1-\tan x\right)\right\} \\ =\frac{1}{2}\left\{\frac{1}{1+\tan x}\left(0+{\sec }^{2}x\right)-\frac{1}{1-\tan x}\left(0-{\sec }^{2}x\right)\right\} \\ =\frac{1}{2}\left\{\frac{{\sec }^{2}x}{1+\tan x}+\frac{{\sec }^{2}x}{1-\tan x}\right\} \\ =\frac{1}{2}{\sec }^{2}x\left\{\frac{1-\tan x+1+\tan x}{1-{\tan }^{2}x}\right\} \\ =\frac{1}{2}{\sec }^{2}x\left(\frac{2}{1-{\tan }^{2}x}\right) \\ =\frac{{\sec }^{2}x}{1-{\tan }^{2}x} \\ =\frac{1+{\tan }^{2}x}{1-{\tan }^{2}x} \\ =\frac{1}{{\displaystyle \frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}}} \\ =\frac{1}{\cos 2x} \\ =\sec 2x \end{gathered}$$