#### Answer

$f'(x) = -cos(cot(x))csc^{2}(x)$

#### Work Step by Step

First write $f(g(x))$ in terms of $u$ and $f(u)$.
Original expression: $y = sin(cot(x))$
$u = g(x) = cot(x)$
$y = f(u) = sin(u)$
Apply the chain rule to find the derivative:
$f'(g(x)) \times g'(x) $ = > $f'(u) \times u'$ => $\frac{dy}{du}sin(u) \times\frac{du}{dx}(cot(x)) $
Derive:
$f'(u) = cos(u)\times-csc^2(x)$
$f'(x) = cos(cot(x))\times-csc^2(x)=-cos(cot(x))csc^{2}(x)$