Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 29

If $\displaystyle f(x) = \frac{1}{\sqrt{x + 2}}$, 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{1}{\sqrt{a + h + 2}} - \frac{1}{\sqrt{a + 2}}}{h}
&& \text{Substitute $f(a + h)$ and $f(a)$}\\
\\
f'(a) &= \lim_{h \to 0} \frac{\sqrt{a + 2} - \sqrt{a + h + 2}}{(h)(\sqrt{a + h + 2})(\sqrt{a + 2})}
&& \text{Get the LCD of the numerator}\\
\\
f'(a) &= \lim_{h \to 0} \frac{\sqrt{a + 2} - \sqrt{a + h + 2}}{(h)(\sqrt{a + h + 2})(\sqrt{a + 2})}
\cdot \frac{\sqrt{a + 2} + \sqrt{a + h + 2}}{\sqrt{a + 2} + \sqrt{a + h + 2}}
&& \text{Multiply both numerator and denominator by $(\sqrt{a + 2} + \sqrt{a + h + 2})$}\\
\\
f'(a) &= \lim_{h \to 0} \frac{a + 2 - ( a + h + 2)}{(h)(\sqrt{a + h + 2})(\sqrt{a + 2})(\sqrt{a + 2} + \sqrt{a + h + 2})}
&& \text{Simplify the equation}\\
\\
f'(a) &= \lim_{h \to 0} \frac{\cancel{a} + \cancel{2} - \cancel{a} - h - \cancel{2}}{(h)(\sqrt{a + h + 2})(\sqrt{a + 2})(\sqrt{a + 2} + \sqrt{a + h + 2})}
&& \text{Combine like terms}\\
\\
f'(a) &= \lim_{h \to 0} \frac{\cancel{-h}}{\cancel{(h)} (\sqrt{a + h + 2}) (\sqrt{a + 2})(\sqrt{a + 2} + \sqrt{a + h + 2})}
&& \text{Cancel out like terms}\\
\\
f'(a) &= \lim_{h \to 0} \left[ \frac{-1}{(\sqrt{a + h + 2})(\sqrt{a + 2})(\sqrt{a + 2} + \sqrt{a + h + 2})}\right] = \frac{-1}{(\sqrt{a + 0 + 2})(\sqrt{a + 2})(\sqrt{a + 2} + \sqrt{a + 0 + 2})}
&& \text{Evaluate the limit}\\
\\
f'(a) &= \frac{-1}{(\sqrt{a + 2})(\sqrt{a + 2})(\sqrt{a + 2} + \sqrt{a + 2})} = \frac{-1}{(a + 2)(2 \sqrt{a + 2})} = \frac{-1}{(a + 2)(2)(a + 2)^{\frac{1}{2}}}
&& \text{Simplify the equation}

\end{aligned}
\end{equation}
$


$\qquad\fbox{$f'(a) = \displaystyle \frac{-1}{2(a + 2)^{\frac{3}{2}}}$} $