Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 30

If $\displaystyle f(x) = \sqrt{3x + 1}$, 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{\sqrt{3(a + h) + 1} - \sqrt{3a + 1}}{h}
&& \text{Substitute $f(a + h)$ and $f(a)$}\\
\\
f'(a) &= \lim_{h \to 0} \frac{\sqrt{3a + 3h + 1} - \sqrt{3a + 1}}{h} \cdot \frac{\sqrt{3a + 3h + 1} + \sqrt{3a + 1}}{\sqrt{3a + 3h + 1} + \sqrt{3a + 1}} && \text{Multiply both numerator and denominator by $(\sqrt{3a + 3h + 1} + \sqrt{3a + 1})$}\\
\\
f'(a) &= \lim_{h \to 0} \frac{3a + 3h + 1 -(3a +1)}{(h)(\sqrt{3a + 3h + 1} + \sqrt{3a + 1})}
&& \text{Simplify the equation}\\
\\
f'(a) &= \lim_{h \to 0} \frac{\cancel{3a} + 3h + \cancel{1} - \cancel{3a} - \cancel{1}}{(h)(\sqrt{3a + 3h + 1} + \sqrt{3a + 1})}
&& \text{Combine like terms}\\
\\
f'(a) &= \lim_{h \to 0} \frac{3 \cancel{h}}{\cancel{(h)}(\sqrt{3a + 3h + 1} + \sqrt{3a + 1})}
&& \text{Cancel out like terms}\\
\\
f'(a) &= \lim_{h \to 0} \frac{3}{\sqrt{3a + 3h + 1} + \sqrt{3a + 1}} = \frac{3}{\sqrt{3a + 3(0) + 1} + \sqrt{3a + 1}} = \frac{3}{\sqrt{3a + 1} + \sqrt{3a + 1}}
&& \text{Evaluate the limit}

\end{aligned}
\end{equation}
$


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

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