Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 114

Differentiate $f(x) = x(3x^3 + 6x - 2)(3x^4 + 7)$

If we simplify the function first before we take the derivative, we get

$
\begin{equation}
\begin{aligned}
f(x) &=(3x^4 + 6x^2 - 2x)(3x^4 +7)\\
\\
&= 9x^8 + 21x^4 + 18x^6 + 42x^2 - 6x^5 - 14x
\end{aligned}
\end{equation}
$


Thus, by applying power rule,

$
\begin{equation}
\begin{aligned}
f'(x) &= \frac{d}{dx} \left[ 9x^8 + 18x^4 - 6x^5 + 21x^4 + 42x^2 - 14x \right]\\
\\
&= 9 \cdot 8 x^{8 - 1} + 18 \cdot 6 x^{6 - 1} - 6 \cdot 5 x^{5 - 1} + 21 \cdot 4 x^{4 - 1} + 42 \cdot 2 x^{2-1} - 14(1)\\
\\
&= 72x^7 + 108x^5 - 30x^4 + 84x^3 + 84x - 14
\end{aligned}
\end{equation}
$