Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 25

If $f(x) = 3-2x + 4x^2$, find $f'(a)$.

Using the definition of the derivative


$
\begin{equation}
\begin{aligned}

\qquad f'(a) &= \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}
&& \\
\\
\qquad f'(a) &= \lim_{h \to 0} \frac{3 - 2 (a + h) + 4 (a + h)^2 - (3 - 2a + 4a^2)}{h}
&& \text{Substitute $f(a + h)$ and $f(a)$}\\
\\
\qquad f'(a) &= \lim_{h \to 0} \frac{3 - 2a - 2h + 4(a^2 + 2ah + h^2) - 3 + 2a - 4a^2}{h}
&& \text{Expand the equation}\\
\\
\qquad f'(a) &= \lim_{h \to 0} \frac{\cancel{3} - \cancel{2a} - 2h + \cancel{4a^2} + 8ah + 4h^2 -\cancel{3} + \cancel{2a} - \cancel{4a^2}}{h}
&& \text{Combine like terms}\\
\\
\qquad f'(a) &= \lim_{h \to 0} \frac{4h^2 + 8ah -2h}{h}
&& \text{Factor the numerator}\\
\\
\qquad f'(a) &= \lim_{h \to 0} \frac{\cancel{h}(4h + 8a - 2)}{\cancel{h}}
&& \text{Cancel out like terms}\\
\\
\qquad f'(a) &= \lim_{h \to 0} (4h + 8a - 2) = 4(0) + 8a - 2
&& \text{Evaluate the limit}

\end{aligned}
\end{equation}
$


$\qquad\fbox{$f'(a) =8 a - 2$} $