$$ \left. \begin{array} { l } { \quad f ( x ) = 5 \sqrt { x } \sin ( x ) } \ { f ^ { \prime } ( x ) = \square } \end{array} ight. $$

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$$ \left. \begin{array} { l } { \quad f ( x ) = 5 \sqrt { x } \sin ( x ) } \ { f ^ { \prime } ( x ) = \square } \end{array} ight. $$

UpStudy AI · Accepted Answer

Given
$$
f(x)=5\sqrt{x}\sin(x),
$$
we differentiate using the product rule. Write $$ u(x)=5\sqrt{x} $$ and $$ v(x)=\sin(x) $$, so that
$$
f(x)=u(x)v(x).
$$

1. Differentiate $$ u(x)=5\sqrt{x} $$. Recall that $$\sqrt{x}=x^{1/2}$$, so
   $$
   u'(x)=5\cdot\frac{1}{2}x^{-1/2}=\frac{5}{2}\frac{1}{\sqrt{x}}.
   $$

2. Differentiate $$ v(x)=\sin(x) $$, which gives
   $$
   v'(x)=\cos(x).
   $$

3. Apply the product rule:
   $$
   f'(x)=u'(x)v(x)+u(x)v'(x)=\frac{5}{2}\frac{1}{\sqrt{x}}\sin(x)+5\sqrt{x}\cos(x).
   $$

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

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