If y = ( log cot x tan x ) ( log tan x cot x ) + tan - 1 4 x 4 - x 2 , then d y d x =

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

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

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

$y^{\text{'}}=0+\frac{d}{dx}\left({\tan }^{-1}\left(\frac{4x}{4-x^2}\right)\right)$

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

$=\frac{1}{1+{\displaystyle \frac{16x^2}{16-8x^2+x^4}}}\frac{\left(4-x^2\right)4-4x\left(-2x\right)}{{\left(4-x^2\right)}^2}$

$=\frac{4}{4+x^2}$