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.