PREVIOUS ANSWERS Differentiate. $$ f(x)=5 \sqrt{x} \sin (x) $$ $$ f^{\prime}(x)= $$

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We start with the function
$$
f(x)=5\sqrt{x}\sin(x)=5x^{\frac{1}{2}}\sin(x).
$$
We will use the product rule, which states that if
$$
h(x)=u(x)v(x),
$$
then
$$
h'(x)=u'(x)v(x)+u(x)v'(x).
$$

Let 
$$
u(x)=x^{\frac{1}{2}} \quad \text{and} \quad v(x)=\sin(x).
$$

First, we differentiate $$ u(x) $$ and $$ v(x) $$:

1. Differentiate $$ u(x)=x^{\frac{1}{2}} $$:
   $$
   u'(x)=\frac{1}{2}x^{-\frac{1}{2}}.
   $$

2. Differentiate $$ v(x)=\sin(x) $$:
   $$
   v'(x)=\cos(x).
   $$

Applying the product rule:
$$
\frac{d}{dx}\left[x^{\frac{1}{2}}\sin(x)\right]=u'(x)v(x)+u(x)v'(x)=\frac{1}{2}x^{-\frac{1}{2}}\sin(x)+x^{\frac{1}{2}}\cos(x).
$$

Since $$ f(x)=5x^{\frac{1}{2}}\sin(x) $$, we multiply the derivative by $$ 5 $$:
$$
f'(x)=5\left(\frac{1}{2}x^{-\frac{1}{2}}\sin(x)+x^{\frac{1}{2}}\cos(x)\right).
$$

Simplify:
$$
f'(x)=\frac{5}{2}x^{-\frac{1}{2}}\sin(x)+5x^{\frac{1}{2}}\cos(x).
$$

Thus, the derivative is:
$$
f'(x)=\frac{5\sin(x)}{2\sqrt{x}}+5\sqrt{x}\cos(x).
$$
$$ f^{\prime}(x) = \frac{5\sin(x)}{2\sqrt{x}} + 5\sqrt{x}\cos(x) $$

PREVIOUS ANSWERS
Differentiate.

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