Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 27

If $\displaystyle f(t) = \frac{2t + 1}{t + 3}$, 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{\displaystyle \frac{2(a + h) + 1}{a+ h + 3} - \left( \frac{2a + 1}{a + 3} \right) }{h}
&& \text{Substitute $f(a + h)$ and $f(a)$}\\
\\
f'(a) &= \lim_{h \to 0} \frac{(2a + 2h + 1)(a + 3) - (a + h + 3)(2a + 1)}{(h)(a + h + 3)(a + 3)}
&& \text{Get the LCD of the numerator and simplify}\\
\\
f'(a) &= \lim_{h \to 0} \frac{2a^2 + 2ah + a + 6a + 6h +3 - (2a^2 + 2ah + 6a + a + h + 3)}{(h)(a + h + 3)(a + 3)}
&& \text{Expand the equation}\\
\\
f'(a) &= \lim_{h \to 0} \frac{\cancel{2a^2} + \cancel{2ah} + \cancel{a} + \cancel{6a} + 6h + \cancel{3} - \cancel{2a^2} - \cancel{2ah} - \cancel{6a} - \cancel{a} - h - \cancel{3}}{(h)(a + h + 3)(a + 3)}
&& \text{Combine like terms}\\
\\
f'(a) &= \lim_{h \to 0} \frac{5\cancel{h}}{\cancel{(h)}(a + h + 3)(a + 3)}
&& \text{Cancel out like terms}\\
\\
f'(a) &= \lim_{h \to 0} \left[\frac{5}{(a + h + 3)(a + 3)} \right] = \frac{5}{(a + 0 + 3)(a + 3)}
&& \text{Evaluate the limit}

\end{aligned}
\end{equation}
$


$\qquad\fbox{$f'(a) = \displaystyle \frac{5}{(a + 3)^2}$} $