#### Answer

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

#### Work Step by Step

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