Answer
$y' = \frac{2(x+1)(x+4)}{(x^2-4)(2x+5)}$
Work Step by Step
Let $f(x) = \frac{x^2-4}{2x+5}$
$f'(x) = \frac{2x(2x+5)-2(x^2-4)}{(2x+5)^2}$
$f'(x) = \frac{4x^2+10x-2x^2+8}{(2x+5)^2}$
$f'(x) = \frac{2x^2+10x+8}{(2x+5)^2}$
$f'(x) = \frac{2(x+1)(x+4)}{(2x+5)^2}$
If $y = ln\vert f(x) \vert,~~$ then $y' = \frac{1}{f(x)}\cdot f'(x)$
$y = ln\vert \frac{x^2-4}{2x+5} \vert$
Then:
$y' = \frac{1}{f(x)}\cdot f'(x) $
$y' = \frac{1}{\frac{x^2-4}{2x+5}}\cdot \frac{2(x+1)(x+4)}{(2x+5)^2}$
$y' = \frac{2x+5}{x^2-4}\cdot \frac{2(x+1)(x+4)}{(2x+5)^2}$
$y' = \frac{1}{x^2-4}\cdot \frac{2(x+1)(x+4)}{(2x+5)}$
$y' = \frac{2(x+1)(x+4)}{(x^2-4)(2x+5)}$
