Question

Find the derivative of $y=\frac{1}{4}\ln \frac{x^2-1}{x^2+1}$

Answer 1

$y=\frac{1}{4}ln\frac{x^2-1}{x^2+1}$

$\Rightarrow y=\frac{1}{4}\left[ln\right(x^2-1\left)\right)-ln(x^2+1)]$

$\Rightarrow \frac{dy}{dx}=\frac{1}{4}\left[\frac{d}{dx}\right(ln(x^2-1))-\frac{d}{dx}(ln(x^2+1)\left)\right]$

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

$=\frac{1}{4}[\frac{2x}{x^2-1}-\frac{2x}{x^2+1}]$

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

$=\frac{2x}{4}\times \frac{2}{(x^4-1)}$

$=\frac{4x}{4(x^4-1)}=\frac{x}{x^4-1}$.