Question

If $f(x)={\log }_{\cot x}\tan x\cdot {\log }_{\tan x}\cot x^{-1}+$ ${\tan }^{-1}\frac{x}{\sqrt{4-x^2}},x\in (-2,2)$, then the value of $2f^{\prime }(0)=$ ____

Answer 1

Answer: 1

Let $a={\log }_{\cot x}\tan x{\log }_{\tan x}\cot x^{-1}$
and $b={\tan }^{-1}\frac{x}{\sqrt{4-x^2}}$
$\Rightarrow f(x)=a+b$
$a={\log }_{\cot x}\tan x{\log }_{\tan x}\cot x^{-1}$
$={\log }_{\cot x}\tan x\frac{1}{{\log }_{\tan x}\cot x}$
$=\frac{\log (\tan x{)}^2}{\log (\cot x)}=\frac{\log (\tan x{)}^2}{\log \frac{1}{\tan x}}$
$=\frac{\log (\tan x{)}^2}{-\log (\tan x)}=1$
$b={\tan }^{-1}\frac{x}{\sqrt{4-x^2}}$
Let $\theta ={\sin }^{-1}\frac{x}{2},-2<x<2$
$\Rightarrow x=2\sin \theta$
$\Rightarrow b={\tan }^{-1}\frac{2\sin \theta }{\sqrt{4-4{\sin }^2\theta }}$
$={\tan }^{-1}\frac{2\sin \theta }{2\cos \theta }$
$={\tan }^{-1}(\tan \theta ),-\frac{\pi }{2}<\theta <\frac{\pi }{2}$
$=\theta$
$={\sin }^{-1}\frac{x}{2}$
$\Rightarrow f(x)=1+{\sin }^{-1}\frac{x}{2}$
$\Rightarrow f^{\prime }(x)=\frac{1}{\sqrt{1-\frac{x^2}{2}}}\cdot \frac{1}{2}$
$\Rightarrow f^{\prime }(x)=\frac{1}{\sqrt{4-x^2}}$
$\Rightarrow f^{\prime }(0)=\frac{1}{\sqrt{4-0}}=\frac{1}{2}\Rightarrow 2f^{\prime }(0)=1$