Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 45

Suppose that $f(x) = 2x^2 - x^3$, find $f'(x), f''(x), f'''(x)$ and $f^4(x)$. Graph $f, f', f''$ and $f'''$ on a common screen. Are the graphs consistent with the geometric interpretations of these derivatives?

Using the definition of derivative


$\begin{equation} \begin{aligned} \qquad f'(x) =& \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} && \\ \\ \qquad f'(x) =& \lim_{h \to 0} \frac{2(x + h)^2 - (x + h)^3 - (2x^2 - x^3)}{h} && \text{Substitute $f(x + h)$ and $f(x)$} \\ \\ \qquad f'(x) =& \lim_{h \to 0} \frac{2(x^2 + 2xh + h^2) - (x^3 + 3x^2 h + 3xh^2 + h^3 ) - 2x^2 + x^3}{h} && \text{Expand the equation} \\ \\ \qquad f'(x) =& \lim_{h \to 0} \frac{\cancel{2x^2} + 4xh + 2h^2 - \cancel{x^3} - 3x^2h - 3xh^2 - h^3 - \cancel{2x^2} + \cancel{x^3}}{h} && \text{Combine like terms} \\ \\ \qquad f'(x) =& \lim_{h \to 0} \frac{4xh + 2h^2 - 3x^2 h - 3xh^2 - h^3}{h} && \text{Factor the numerator} \\ \\ \qquad f'(x) =& \lim_{h \to 0} \frac{\cancel{h}(4x + 2h - 3x^2 - 3xh - h^2)}{\cancel{h}} && \text{Cancel out like terms} \\ \\ f'(x) =& \lim_{h \to 0} (4x + 2h - 3x^2 - 3xh - h^2) = 4x + 2(0) - 3x^2 - 3x(0) - (0)^2 = 4x + 0 - 3x^2 - 0 - 0 && \text{Evaluate the limit} \\ \\ f'(x) =& 4x - 3x^2 && \end{aligned} \end{equation}$


Using the 2nd derivative of the definition


$\begin{equation} \begin{aligned} \qquad f''(x) =& \lim_{h \to 0} \frac{f'(x + h) = f'(x)}{h} && \\ \\ \qquad f''(x) =& \lim_{h \to 0} \frac{4(x + h) - 3 (x + h)^2 - (4x - 3x^2)}{h} && \text{Substitute $f'(x + h)$ and $f'(x)$} \\ \\ \qquad f''(x) =& \lim_{h \to 0} \frac{4x + 4h - 3 (x^2 + 2xh + h^2) - 4x + 3x^2}{h} && \text{Expand the equation} \\ \\ \qquad f''(x) =& \lim_{h \to 0} \frac{\cancel{4x} + 4h - \cancel{3x^2} - 6xh - 3h^2 - \cancel{4x} + \cancel{3x^2}}{h} && \text{Combine like terms} \\ \\ \qquad f''(x) =& \lim_{h \to 0} \frac{4h - 6xh - 3h^2}{h} && \text{Factor the numerator} \\ \\ \qquad f''(x) =& \lim_{h \to 0} \frac{\cancel{h}(4 - 6x - 3h)}{\cancel{h}} && \text{Cancel out like terms} \\ \\ \qquad f''(x) =& \lim_{h \to 0} (4 - 6x - 3h) = 4 - 6x - 3(0) = 4 - 6x - 0 && \text{Evaluate the limit} \\ \\ \qquad f''(x) =& 4 - 6x \end{aligned} \end{equation}$


Using the 3rd derivative of the definition


$\begin{equation} \begin{aligned} \qquad f'''(x) =& \lim_{h \to 0} \frac{f''(x + h) = f''(x)}{h} && \\ \\ \qquad f'''(x) =& \lim_{h \to 0} \frac{4 - 6 (x + h) - (4 - 6x)}{h} && \text{Substitute $f''(x + h)$ and $f''(x)$} \\ \\ \qquad f'''(x) =& \lim_{h \to 0} \frac{\cancel{4} - \cancel{6x} - 6h - \cancel{4} + \cancel{6x}}{h} && \text{Expand the equation and combine like terms} \\ \\\qquad f'''(x) =& \lim_{h \to 0}\frac{-6\cancel{h}}{\cancel{h}} && \text{Cancel out like terms} \\ \\ \qquad f'''(x) =& -6 && \end{aligned} \end{equation}$


Using the 4th derivative of the definition


$\begin{equation} \begin{aligned} \qquad f^4(x) =& \lim_{h \to 0} \frac{f'''(x + h) = f'''(x)}{h} && \\ \\ \text{If $f'''$ is constant, then $f'''(x + h) = f'''(x)$} \\ \\ \qquad f^4(x) =& \lim_{h \to 0} \frac{-6 - (-6)}{h} && \text{Substitute $f'''(x + h)$ and $f'''(x)$} \\ \\ \qquad f^4(x) =& \lim_{h \to 0} \frac{-6 + 6}{h} = \lim_{h \to 0} \frac{0}{h} && \text{Simplify the equation} \\ \\ \qquad f^4(x) =& 0 \end{aligned} \end{equation}$




Graph $f, f', f''$ and $f'''$