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 $\frac{-4x}{1-x^4}$

$y=\log \left(\frac{1-x^2}{1+x^2}\right)$
$\Rightarrow \frac{dy}{dx}=\frac{1}{\frac{1-x^2}{1+x^2}}\cdot \frac{-2x(1+x^2)-2x(1-x^2)}{(1+x^2{)}^2}$
$\Rightarrow \frac{dy}{dx}=\frac{1+x^2}{1-x^2}\cdot \frac{-4x}{(1+x^2{)}^2}$
$\Rightarrow \frac{dy}{dx}=\frac{-4x}{1-x^4}$