Помогите решить производную

Решите задачу:

$\frac{d}{dx}[\arccos(\sin(x))^{ \frac{1}{3}}]$
$\frac{1}{3} \arccos(\sin(x))^{ \frac{1}{3} -1} \frac{d}{dx}[\arccos(\sin(x))]$
$\frac{1}{3} \arccos(\sin(x))^{ \frac{1}{3} + \frac{-1}{3} \frac{3}{3} } \frac{d}{dx} [\arccos (\sin(x))]$
$\frac{1}{3} \arccos(\sin(x))^{ \frac{1}{3} + \frac{-1*3}{3} } \frac{d}{dx} [\arccos(\sin(x))]$
$\frac{\arccos(\sin(x))^{- \frac{2}{3}} }{3} (- \frac{1}{ \sqrt{1-\sin^2(x)}} \frac{d}{dx} [\sin(x)] }$
$- \frac{\arccos(\sin(x))^{- \frac{2}{3}} }{3 \sqrt{1-\sin^2(x)} } \cos(x)$
$- \frac{\cos(x)\arccos(\sin(x))^{- \frac{2}{3}}}{3 \sqrt{1-\sin^2(x)} }$
$- \frac{\cos(x) \frac{1}{\arccos(\sin(x))^{- \frac{2}{3}}} }{3 \sqrt{1-\sin^2(x)} }$
$- \frac{\cos(x)}{3\arccos(\sin(x))^ \frac{2}{3} \sqrt{1-\sin^2(x)} }$