If $y=\log \left(\frac{1-x^2}{1+x^2}\right)$, then $\frac{d y}{d x}$ is :
Answer & explanation
Correct answer: option 3
$y=\log \left(1-x^2\right)-\log \left(1+x^2\right)$
$\Rightarrow \frac{d y}{d x}=\frac{-2 x}{1-x^2}-\frac{2 x}{1+x^2}=\frac{-4 x}{1-x^4}$
Hence (1) is correct answer.