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 $f'''$