Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 43

Find $f'(x)$ and $f''(x)$ on the function $f(x) = 1 + 4x - x^2$ using the definition of a derivative. Then graph $f, f'$ and $f''$ on a common screen and check to see if your answers are reasonable.

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{1 + 4 (x + h) - (1 + 4x - x^2)}{h}
&& \text{Substitute $f(x + h)$ and $f(x)$}
\\
\\
\qquad f'(x) =& \lim_{h \to 0} \frac{1 + 4x + 4h - (x^2 + 2xh + h^2) - 1 - 4x + x^2}{h}
&& \text{Expand the equation}
\\
\\
\qquad f'(x) =& \lim_{h \to 0} \frac{\cancel{1} + \cancel{4x} + 4h - \cancel{x^2} - 2xh - h^2 - \cancel{1} - \cancel{4x} + \cancel{x^2}}{h}
&& \text{Combine like terms}
\\
\\
\qquad f'(x) =& \lim_{h \to 0} \frac{4h - 2xh - h^2}{h}
&& \text{Factor the numerator}
\\
\\
\qquad f'(x) =& \lim_{h \to 0} \frac{\cancel{h}(4 - 2x - h)}{\cancel{h}}
&& \text{Cancel out like terms}
\\
\\
f'(x) =& \lim_{h \to 0} (4 - 2x - h) \quad = 4 - 2x - 0
&& \text{Evaluate the limit}
\\
\\
f'(x) =& 4 - 2x
&&

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


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