Answer
$f'(x) = g'(tan(\sqrt x)) \frac {sec^{2} (\sqrt x)} {2\sqrt x}$
Work Step by Step
$f(x) = g(tan(\sqrt x))$
Use chain rule here
$f'(x) = \frac{d}{dx} g(tan(\sqrt x))$
$f'(x) = g'(tan(\sqrt x)) \frac{d}{dx} tan(\sqrt x))$
$f'(x) = g'(tan(\sqrt x)) sec^{2} (\sqrt x)\frac{d}{dx} \sqrt x$
$f'(x) = g'(tan(\sqrt x)) sec^{2} (\sqrt x)\frac{1}{2\sqrt x}$
$f'(x) = g'(tan(\sqrt x)) \frac {sec^{2} (\sqrt x)} {2\sqrt x}$
