Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 12

Take the derivative of $G(t) = (2t + 3 \sqrt{t} +5)(\sqrt{t} + 4)$: first, use the Product Rule; then,
by multiplying the expression before differentiating. Compare your results as a check.
By using Product Rule,

$\begin{equation} \begin{aligned} G'(t) = \frac{d}{dx} \left[ (2t + 3\sqrt{t} +5)(\sqrt{t} +4) \right] &= (2t + 3\sqrt{t} + 5) \cdot \frac{d}{dt} (\sqrt{t} + 4 ) + (\sqrt{t} + 4) \cdot \frac{d}{dt} (2t + 3 \sqrt{t} +5 )\\ \\ &= (2t + 3\sqrt{t} +5) \left( \frac{1}{2\sqrt{t}} \right) + (\sqrt{t} + 4) \left( 2 + \frac{3}{2\sqrt{t}} \right)\\ \\ &= \left[ \frac{t}{\sqrt{t}} + \frac{3}{2} + \frac{5}{2\sqrt{t}} \right] + \left[ 2\sqrt{t} + \frac{3}{2} + 8 + \frac{6}{\sqrt{t}} \right]\\ \\ &= \sqrt{t} + \frac{3}{2} + \frac{5}{2\sqrt{t}} + 2 \sqrt{t} + \frac{3}{2} + 8 \frac{6}{\sqrt{t}}\\ \\ &= 3 \sqrt{t} + \frac{17}{2 \sqrt{t}} + 11 \end{aligned} \end{equation}$


By multiplying the expression first,

$\begin{equation} \begin{aligned} G(t) &= (2t + 3 \sqrt{t} + 5) (\sqrt{t} + 4)\\ \\ &= 2t^{\frac{3}{2}} + 3t + 5t^{\frac{1}{2}} + 8t + 12t^{\frac{1}{2}} + 20\\ \\ &= 2t^{\frac{3}{2}} + 17t^{\frac{1}{2}} + 11t + 20\\ \\ G'(t) &= \frac{d}{dt} \left[ 2t^{\frac{3}{2}} + 17t^{\frac{1}{2}} + 11t + 20 \right]\\ \\ &= 2 \cdot \frac{3}{2} + t^{\frac{3}{2} - 1} + 17 \cdot \frac{1}{2}t^{\frac{1}{2} - 1} + 11(1) + 0 \\ \\ &= 3t^{\frac{1}{2}} + \frac{17}{2} t^{-\frac{1}{2}} + 11\\ \\ &= 3 \sqrt{t} + \frac{17}{2 \sqrt{t}} + 11 \end{aligned} \end{equation}$


Both results agree.