Question

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

• 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)

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