Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 19

Find the derivative of $\displaystyle f(t) = 5t - 9t^2$ using the definition and the domain of its derivative.

Using the definition of derivative


$
\begin{equation}
\begin{aligned}

\qquad f'(t) &= \lim_{h \to 0} \frac{f(t + h) - f(t)}{h}
&&
\\
\\
\qquad f'(t) &= \lim_{h \to 0} \frac{5(t + h) - 9(t + h)^2 - (5t - 9t^2)}{h}
&& \text{Substitute $f(t + h)$ and $f(t)$}
\\
\\
\qquad f'(t) &= \lim_{h \to 0} \frac{\cancel{5t} + 5h - \cancel{9t^2} - 18th - 9h^2 - \cancel{5t} + \cancel{9t^2}}{h}
&& \text{Expand and combine like terms}
\\
\\
\qquad f'(t) &= \lim_{h \to 0} \frac{5h - 18th - 9h^2}{h}
&& \text{Factor the numerator}
\\
\\
\qquad f'(t) &= \lim_{h \to 0} \frac{\cancel{h}(5 - 18t - 9h)}{\cancel{h}}
&& \text{Cancel out like terms}
\\
\\
\qquad f'(t) &= \lim_{h \to 0} 5 - 18t - 9h = 5 - 18t - 9(0)
&& \text{Evaluate the limit}

\end{aligned}
\end{equation}
$


$\qquad \fbox{$f'(t) = 5 - 18t$}$

$f(t)$ is a polynomial function while $f'(t)$ is a linear function. Both of the functions are continuous in every number. Therefore, their domain is $(-\infty, \infty)$