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.