Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 10

Take the derivative of $f(x) = (2x + 5)(3x^2 -4x + 1)$: 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}
f'(x) = \frac{d}{dx} \left[ (2x +5)(3x^2 - 4x + 1) \right] &= (2x + 5) \cdot \frac{d}{dx} (3x^2 - 4x + 1) + (3x^2 - 4x + 1) \cdot \frac{d}{dx} (2x + 5)\\
\\
&= (2x + 5)(6x - 4) + (3x^2 - 4x + 1)(2)\\
\\
&= \left[ 12x^2 - 8x + 30x - 20 \right] + \left[ 6x^2 - 8x + 2 \right]\\
\\
&= 18x^2 + 14x - 18
\end{aligned}
\end{equation}
$


By multiplying the expression first,

$
\begin{equation}
\begin{aligned}
f(x) = (2x + 5)(3x^2 - 4x + 1) &= 6x^3 - 8x^2 + 2x + 15x^2 - 20x + 5 \\
\\
&= 6x^3 + 7x^2 - 18x + 5\\
\\
f'(x) = \frac{d}{dx} \left[ 6x^3 + 7x^2 - 18x + 5 \right] &= 18x^2 + 14x - 18
\end{aligned}
\end{equation}
$


Both results agree.