Question

Find $\frac{dy}{dx}$ for $y=\log \left(\frac{1-x^2}{1+x^2}\right)$.

Answer 1

Given,

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

or, $y=\log (1-x^2)-\log (1+x^2)$

Now, differentiating both sides with respect to $x$ we get,

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

or, $\frac{dy}{dx}=\frac{-4x}{1-x^4}$.