Question

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

Answer 1

Given,

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

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

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

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

or, $\frac{dy}{dx}=\frac{2}{1-x^2}$.