Question

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

Answer 1

Given,

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

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

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

or, $\frac{dy}{dx}=\frac{2x}{1+x^2}+\frac{\left(2x\right)}{1-x^2}$

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