Question

If $f\left(x\right)=\left({\log }_{\cot x}\tan x\right)({\log }_{\tan x}\cot x{)}^{-1}+{\tan }^{-1}\left(\frac{4x}{4-x^2}\right)$, then $f^{\prime }\left(2\right)$ is equal to

• A. $2$
• B. $0$
• C. $\frac{1}{2}$
• D. $-2$

Answer 1

Answer: (C)

$\frac{1}{2}$

$f\left(x\right)=\frac{\log \tan x}{\log \cot x}.\frac{\log \tan x}{\log \cot x}+{\tan }^{-1}\left(\frac{4x}{4-x^2}\right)$

$={\tan }^{-1}\left(\frac{4x}{4-x^2}\right)+{\left[\frac{\log \tan x}{\log (tanx{)}^{-1}}\right]}^2$

$=1+{\tan }^{-1}\left(\frac{4x}{4-x^2}\right)=1+2{\tan }^{-1}\left(\frac{x}{2}\right)$

$f^{\prime }\left(x\right)=\frac{4x}{4+x^2}$

$\therefore f^{\prime }\left(2\right)=\frac{1}{2}$