Question

If $y={\log }_{\cot x}\tan x{\log }_{\tan x}\cot x+{\tan }^{-1}\frac{4x}{4-x^2}$ then $\frac{dy}{dx}=$

• A. $\frac{1}{4+x^2}$
• B. $\frac{4}{4+x^2}$
• C. $\frac{1}{4-x^2}$
• D. $\frac{4}{4-x^2}$

Answer 1

Answer: (B)

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