Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 57

Find the 1st and 2nd derivatives of $f(x) = x^4 - 3x^3 + 16x$


$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{d}{dx} (x^4) - 3 \frac{d}{dx} (x^3) + 16 \frac{d}{dx} (x)
&& \text{Derive each term}
\\
\\
f'(x) =& 4x^3 - 3 (3x^2) + 16 (1)
&& \text{Simplify the equation}
\\
\\
f'(x) =& 4x^3 - 9x^2 + 16
&& \text{1st derivative of $f(x)$}
\\
\\
\\
\\
f''(x) =& 4 \frac{d}{dx} (x^3) - 9 \frac{d}{dx} (x^2) + \frac{d}{dx} (16)
&& \text{Derive each term to get 2nd derivative}
\\
\\
f''(x) =& 4(3x^2) - 9 (2x) + 0
&& \text{Simplify the equation}
\\
\\
f''(x) =& 12x^2 - 18x
&& \text{2nd derivative $f(x)$}

\end{aligned}
\end{equation}
$