$y=\log _{e}\left(\frac{1-x^2}{1+x^2}\right)$, then $\frac{d y}{d x}$ is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{-4 x}{1-x^4}$
$y=\log _{e}\left(\frac{1-x^2}{1+x^2}\right)$
$⇒\frac{dy}{dx}=\frac{1+x^2}{1-x^2}\left(\frac{(-2x)(1+x^2)-2x(1-x^2)}{(1+x^2)^2}\right)$
$=\frac{1+x^2}{1-x^2}×\frac{-4x}{(1+x^2)^2}$
$=\frac{-4 x}{1-x^4}$