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}$