Calculus and Its Applications, Chapter 1, 1.8, Section 1.8, Problem 22

Determine the $f'(x)$ of the function $\displaystyle f(x) = (3x^2 + 2x + 1)^{10}$
By using Chain Rule, we get

$
\begin{equation}
\begin{aligned}
f'(x) = \frac{d}{dx} \left( 3x^2 + 2x + 1 \right)^5 &= 5 (3x^2 + 2x + 1)^{5 - 1} \cdot \frac{d}{dx} (3x^2 + 2x + 1)\\
\\
&= 5 (3x^2 + 2x + 1)^4 (6x + 2)\\
\\
&= 5 (6x + 2)(3x^2 + 2x + 1)^4
\end{aligned}
\end{equation}
$


Then, by using Product and Chain Rule

$
\begin{equation}
\begin{aligned}
f''(x) &= 5 \left[ (6x + 2) \cdot \frac{d}{dx} (3x^2 + 2x + 1)^4 + (3x^2 + 2x + 1)^4 \cdot \frac{d}{dx} (6x + 2 ) \right]\\
\\
&= 5 \left[ (6x + 2) \cdot 4(3x^2 + 2x + 1)^{4 - 1} \cdot \frac{d}{dx}(3x^2 + 2x + 1) + (3x^2 + 2x + 1)(6) \right]\\
\\
&= 5\left[ (6x + 2) \cdot 4(3x^2 + 2x + 1)^3 (6x +2) + (3x^2 + 2x + 1)(6) \right]\\
\\
&= 5 \left[ 4 (6x + 2)^2 (3x^2 + 2x + 1)^3 + 6 (3x^2 + 2x + 1) \right]\\
\\
&= 10(3x^2 + 2x + 1) \left[ 2(6x + 2)^2 (3x^2 + 2x + 1)^2 + 3 \right]
\end{aligned}
\end{equation}
$