1. ## differential trig problem

HI, can someone please shed some light on this for me. I spent last night on it without luck. Thank you

show that
d/dx(ln(sqrt(2)-cos x)/(sqrt(2)+cos x) = (2sqrt(2) sin x)/(1+sin^2 x)

2. Carried out step by step, we derive

$f'(x) = \frac{d}{dx}\left(\ln\,\frac{\sqrt{2}-\cos\,x}{\sqrt{2}+\cos\,x}\right)$
$= \frac{d}{dx}(\ln\,(\sqrt{2}-\cos\,x)-\ln\,(\sqrt{2}+\cos\,x))$
$= \frac{d}{dx}(\ln\,(\sqrt{2}-\cos\,x))-\frac{d}{dx}(\ln\,(\sqrt{2}+\cos\,x))$
$= \frac{1}{\sqrt{2}-\cos\,x}\cdot \sin\,x-\frac{1}{\sqrt{2}+\cos\,x}\cdot(-\sin\,x)$
$= \frac{\sin\,x}{\sqrt{2}-\cos\,x}+\frac{\sin\,x}{\sqrt{2}+\cos\,x}$
$= \frac{\sin\,x(\sqrt{2}+\cos\,x)+\sin\,x(\sqrt{2}-\cos\,x)}{2-\cos^2 x}$
$= \frac{\sqrt{2}\sin\,x+\sin\,x\,\cos\,x+\sqrt{2}\si n\,x-\sin\,x\,\cos\,x}{1+1-\cos^2 x}$
$=\frac{2\sqrt{2}\sin\,x}{1+\sin^2 x}.$

3. thank heaps for the effort you put into that. I really appreciate it.