Practicing Success
Practicing Success
CUET Preparation Today
$y=\log _{e}\left(\frac{1-x^2}{1+x^2}\right)$, then $\frac{d y}{d x}$ is equal to: |
$\frac{4 x^3}{1-x^4}$ $\frac{-4 x}{1-x^4}$ $\frac{1}{4-x^4}$ $\frac{-4 x^3}{1-x^4}$ |
$\frac{-4 x}{1-x^4}$ |
The correct answer is Option (2) → $\frac{-4 x}{1-x^4}$ |
