## Calculus 8th Edition

$f'(x) = \frac{cos(x)}{2\sqrt sinx}$
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: $f'(g(x)) \times g'(x)$ = > $f'(u) \times u'$ => $\frac{dy}{du} u^\frac{1}{2} \times\frac{du}{dx}(sin(x))$ $f(u) = \frac{1}{2}u^{-\frac{1}{2} } \times cos(x)$ $f'(u) = \frac{1}{2\sqrt u}\times cos(x)$ $f'(x) = \frac{1}{2\sqrt sinx}\times cos(x)$ $f'(x) = \frac{cos(x)}{2\sqrt sinx}$