Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 22

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

Using the definition of derivative


$
\qquad
\begin{equation}
\begin{aligned}
f'(x) &= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\\

\\
f'(x) &= \lim_{h \to 0} \frac{x + h + \sqrt{x + h} - (x + \sqrt{x})}{h}
&& \text{Substitute } f(x + h) \text{ and } f(x) \\

\\
f'(x) &= \lim_{h \to 0} \frac{\cancel{x} + h \sqrt{x + h} - \cancel{x} - \sqrt{x}}{h}
&& \text{Combine like terms}\\

\\
f'(x) &= \lim_{h \to 0} \frac{\cancel{h}}{\cancel{h}} + \frac{\sqrt{x + h} - \sqrt{x}}{h}
&& \text{Cancel out like terms}\\

\\
f'(x) &= \lim_{h \to 0} 1 + \frac{\sqrt{x + h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}
&& \text{Multiply both numerator and denominator by } \sqrt{x + h} + \sqrt{x}\\

\\
f'(x) &= \lim_{h \to 0} 1 + \frac{x + h - x}{(h)(\sqrt{x + h} + \sqrt{x})}
&& \text{Combine and cancel out like terms}\\

\\
f'(t) &= \lim_{h \to 0} 1 + \frac{1}{\sqrt{x + h} + \sqrt{x}} = 1 + \frac{1}{\sqrt{x + 0} + \sqrt{x}}
&& \text{Evaluate the limit}

\end{aligned}
\end{equation}
$


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

Both functions involves square root that are continuous for $x \geq 0$. However, $\sqrt{x}$ is placed in the denominator of $f'(x)$ that's why is not included in its domain. Therefore,

The domain of $f(x) = x + \sqrt{x}$ is $[0, \infty)$

The domain of $f'(x) = \displaystyle 1 + \frac{1}{2 \sqrt{x}}$ is $(0, \infty)$