Question

If $y=\log \left(\frac{1-x^2}{1+x^2}\right)$, then $\frac{dy}{dx}$ is equal to ___

• A. $\frac{-4x}{1-x^4}$
• B. $\frac{4x^3}{1-x^4}$
• C. $\frac{1}{4-x^4}$
• D. $\frac{-4x^3}{1-x^4}$

Answer 1

The correct option is A $\frac{-4x}{1-x^4}$

We have, $y=\log \left(\frac{1-x^2}{1+x^2}\right)$

$\Rightarrow y=\ln (1-x^2)-\ln (1+x^2)$, using $\log \frac{a}{b}=\log a-\log b$

Now differentiate

$\Rightarrow \frac{dy}{dx}=\frac{-2x}{1-x^2}-\frac{2x}{1+x^2}=-2x(\frac{1}{1-x^2}+\frac{1}{1+x^2})$

$=-2x\left(\frac{1+x^2+1-x^2}{(1-x^2)(1+x^2)}\right)=\frac{-4x}{1-x^4}$