Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 28

If $\displaystyle f(x) = \frac{x^2 + 1}{x - 2}$, 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{(a + h)^2 + 1}{a + h - 2} - \frac{a^2 + 1}{a - 2}}{h}
&& \text{Substitute $f(a + h)$ and $f(a)$}\\
\\
f'(a) &= \lim_{h \to 0} \frac{(a - 2)(a^2 + 2ah + h^2 + 1) - (a^2 + 1)(a + h - 2)}{(h)(a + h -2)(a - 2)}
&& \text{Get the LCD of the numerator and simplify}\\
\\
f'(a) &= \lim_{h \to 0} \frac{a^3 + 2a^2 h + ah^2 +a - 2a^2 - 4ah - 2h^2 - 2 - a^3 - a^2 h +2a^2 - a - h + 2}{(h)(a + h -2)(a - 2)}
&& \text{Expand the equation}\\
\\
f'(a) &= \lim_{h \to 0} \frac{\cancel{a^3} + 2a^2 h + ah^2 + \cancel{a} - \cancel{2a^2} - 4ah - 2h^2 - \cancel{2} - \cancel{a^3} - a^2 h + \cancel{2a^2} - \cancel{a} - h + \cancel{ 2}}{(h)(a + h -2)(a - 2)}
&& \text{Combine like terms}\\
\\
f'(a) &= \lim_{h \to 0} \frac{a^2h + ah^2 - 4ah - 2h^2 - h}{(h)(a + h -2)(a - 2)}
&& \text{Factor the numerator}\\
\\
f'(a) &= \lim_{h \to 0} \frac{\cancel{h}(a^2 + ah - 4 - 2h - 1)}{\cancel{(h)}(a + h - 2)(a - 2)}
&& \text{Cancel out like terms}\\
\\
f'(a) &= \lim_{h \to 0} \left[\frac{a^2 + ah - 4a - 2h - 1}{(a + h - 2)(a - 2)} \right] = \frac{a^2 + a(0) - 4a - 2(0) - 1}{(a + 0 - 2)(a - 2)}
&& \text{Evaluate the limit}

\end{aligned}
\end{equation}
$


$\qquad\fbox{$f'(a) = \displaystyle \frac{a^2 - 4a - 1}{(a - 2)^2}$} $

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