Express the composite function $y = \sqrt{\sin x}$ in the form $f(g(x))$. [Identify the inner function $u=g(x)$ and the outer function $y = f(u)$.] Then find the derivative $\displaystyle \frac{dy}{dx}$
Let $y = f(g(x))$ where $u = g(x) = \sin x$ and $y = f(u) = \sqrt{u}$
Then,
$
\begin{equation}
\begin{aligned}
y' &= \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}\\
\\
y' &= \frac{d}{du}(\sqrt{u}) \cdot \frac{d}{dx} ( \sin x)\\
\\
y' &= \left( \frac{1}{2\sqrt{u}} \right) (\cos x) && \text{Substitute value of }u \text{ and simplify}\\
\\
y' &= \frac{\cos x}{2\sqrt{\sin x}}
\end{aligned}
\end{equation}
$
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