Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 26

If $f(t) = t^4 - 5t$, find $f'(a)$.

Using the definition of the derivative


$
\begin{equation}
\begin{aligned}

f'(a) &= \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}
&& \\
\\
f'(a) &= \lim_{h \to 0} \frac{(a + h)^4 - 5 (a + h) - (a^4 - 5a)}{h}
&& \text{Substitute $f(a + h)$ and $f(a)$}\\
\\
f'(a) &= \lim_{h \to 0} \frac{a^4 + 4a^3 h + 6a^2 h^2 + 4ah^3 + h^4 - 5a - 5h -a^4 + 5a}{h}
&& \text{Expand the equation}\\
\\
f'(a) &= \lim_{h \to 0} \frac{\cancel{a^4} + 4a^3 h + 6a^2 h^2 + 4ah^3 + h^4 - \cancel{5a} - 5h -\cancel{a^4} + \cancel{5a}}{h}
&& \text{Combine like terms}\\
\\
f'(a) &= \lim_{h \to 0} \frac{4a^3 h + 6a^2 h^2 + 4ah^3 + h^4 - 5h}{h}
&& \text{Factor the numerator}\\
\\
f'(a) &= \lim_{h \to 0} \frac{\cancel{h}(4a^3 + 6a^2 h + 4ah^2 +h^3 - 5)}{\cancel{h}}
&& \text{Cancel out like terms}\\
\\
f'(a) &= \lim_{h \to 0} (4a^3 + 6a^2 h + 4ah^2 + h^3 - 5) = 4a^3 + 6a^2 (0) + 4a(0)^2 + (0)^3 - 5
&& \text{Evaluate the limit}

\end{aligned}
\end{equation}
$


$\qquad\fbox{$f'(a) = 4a^3 - 5$} $