$f(x) = \frac{wx}{x^2+b(w+b)}$
$f(\frac{g(x)}{h(x)})=\frac{h(x)g'(x)-g(x)h'(x)}{h(x)^2}$
$\therefore f'(x) = \frac{(x^{2}+b(w+b)) \times w - wx \times 2x}{(x^2+b(w+b))^{2}}$
$\therefore f'(x) = \frac{wx^{2}+wb(w+b) - 2wx^{2}}{(x^2+b(w+b))^{2}}$
$\therefore f'(x) = \frac{w()}{}$
dont worry fellas i got the answer :)
