Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 8

Take the derivative of $y = (4 \sqrt{x} + 3 )x^3$: first, use the Product Rule; then,
by multiplying the expression before differentiating. Compare your results as a check.
By using Product Rule,

$
\begin{equation}
\begin{aligned}
y' = \frac{d}{dx} \left[ (4\sqrt{x} + 3) x^3 \right] &= (4\sqrt{x} + 3) \cdot \frac{d}{dx} (x^3) + x^3 \cdot \frac{d}{dx} (4 \sqrt{x} + 3)\\
\\
&= (4 \sqrt{x} + 3)(3x^2) + x^3 \left( \frac{4}{2\sqrt{x}} \right)\\
\\
&= \left( 4x^{\frac{1}{2}} + 3 \right) (3x^2) + x^3 \left( \frac{2}{x^{\frac{1}{2}}} \right)\\
\\
&= 12x^{\frac{5}{2}} + 9x^2 + 2x^{\frac{5}{2}}\\
\\
&= 14x^{\frac{5}{2}} + 9x^2
\end{aligned}
\end{equation}
$


By multiplying the expression first,

$
\begin{equation}
\begin{aligned}
y = (4 \sqrt{x} + 3)x^3 = \left( 4x^{\frac{1}{2}} + 3 \right) x^3 = 4x^{\frac{7}{2}} + 3x^3 \\
\\
y' &= \frac{d}{dx} \left[ 4x^{\frac{7}{2}} + 3x^3 \right] = 4 \cdot \frac{7}{2} x^{\frac{7}{2} - 1} + 3 \cdot 3 x^{3 - 1}\\
\\
&= 14x^{\frac{5}{2}} + 9x^2 \text{ or } 14\sqrt{x^5} + 9x^2
\end{aligned}
\end{equation}
$


Both results agree.