#### Answer

$f'(x) = \frac{cos(x)}{2\sqrt sinx} $

#### Work Step by Step

First write $f(g(x))$ in terms of $u$ and $f(u)$.
Original expression: $y = \sqrt sin(x)$
$u = g(x) = sin(x)$
$y = f(u) = \sqrt u$
Rewrite $\sqrt u$:
$\sqrt u$ => $u^\frac{1}{2}$
Apply the chain rule to find the derivative:
$f'(g(x)) \times g'(x) $ = > $f'(u) \times u'$ => $\frac{dy}{du} u^\frac{1}{2} \times\frac{du}{dx}(sin(x)) $
$f(u) = \frac{1}{2}u^{-\frac{1}{2} } \times cos(x)$
$f'(u) = \frac{1}{2\sqrt u}\times cos(x)$
$f'(x) = \frac{1}{2\sqrt sinx}\times cos(x)$
$f'(x) = \frac{cos(x)}{2\sqrt sinx} $