Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 23

Find the derivative of $\displaystyle g(x) = \sqrt{1 + 2x}$ using the definition and the domain of its derivative.

Using the definition of derivative


$
\begin{equation}
\begin{aligned}

\qquad f'(x) &= \lim_{h \to 0} \frac{g(x + h) - g(x)}{h}
&&
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{\sqrt{1 + 2(x + h)} - \sqrt{1 + 2x}}{h}
&& \text{Substitute $g(x + h)$ and $g(x)$}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{\sqrt{1 + 2x + 2h} - \sqrt{1 + 2x}}{h} \cdot \frac{\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x}}{\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x}}
&& \text{Multiply both numerator and denominator by $\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x}$}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{\cancel{1} + \cancel{2x} + 2h - \cancel{1} - \cancel{2x}}{(h)(\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x})}
&& \text{Combine like terms}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{2\cancel{h}}{\cancel{(h)}(\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x})}
&& \text{Cancel out like terms}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{2}{(\sqrt{1 + 2x + 2h} + \sqrt{1 + 2x})} = \frac{2}{(\sqrt{1 + 2x + 2(0)} + \sqrt{1 + 2x})}
&& \text{Evaluate the limit}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{\cancel{2}}{\cancel{2} \sqrt{1 + 2x}}
&& \text{Cancel out like terms}
\\
\\
\end{aligned}
\end{equation}
$


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

Both functions involves square root that are continuous for $1 + 2x \geq 0$.

$\displaystyle \begin{array}{cc}
1 + 2x & \geq 0 \\
x & \geq \frac{-1}{2}
\end{array} $

However, $\sqrt{1 + 2x}$ is placed in the denominator of $g'(x)$ that's why $\displaystyle \frac{-1}{2}$ is not included in its domain. Therefore,

The domain of $g(x) = \sqrt{1 + 2x}$ is $\displaystyle \left[ \frac{-1}{2}, \infty \right)$

The domain of $g'(x) = \displaystyle \frac{1}{\sqrt{1 + 2x}}$ is $\displaystyle \left( \frac{-1}{2}, \infty \right)$