# How do you use the definition of a derivative to find the derivative of f(x) = x + sqrtx?

Feb 4, 2016

$f ' \left(x\right) = 1 + \frac{1}{2 \sqrt{x}}$

#### Explanation:

The definition of a derivative is

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

$= {\lim}_{h \to 0} \frac{\left(x + h + \sqrt{x + h}\right) - \left(x + \sqrt{x}\right)}{h}$

$= {\lim}_{h \to 0} \frac{\cancel{x} + h + \sqrt{x + h} - \cancel{x} - \sqrt{x}}{h}$

$= {\lim}_{h \to 0} \frac{h + \sqrt{x + h} - \sqrt{x}}{h}$

$= {\lim}_{h \to 0} \left(\frac{h}{h} + \frac{\sqrt{x + h} - \sqrt{x}}{h}\right)$

$= {\lim}_{h \to 0} 1 + {\lim}_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$

... expand the fraction so that you can use the formula $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$ afterwards...

$= 1 + {\lim}_{h \to 0} \frac{\left(\sqrt{x + h} - \sqrt{x}\right) \cdot \textcolor{b l u e}{\left(\sqrt{x + h} + \sqrt{x}\right)}}{h \cdot \textcolor{b l u e}{\left(\sqrt{x + h} + \sqrt{x}\right)}}$

$= 1 + {\lim}_{h \to 0} \text{ } \frac{{\left(\sqrt{x + h}\right)}^{2} - {\left(\sqrt{x}\right)}^{2}}{h \cdot \left(\sqrt{x + h} + \sqrt{x}\right)}$

$= 1 + {\lim}_{h \to 0} \text{ } \frac{x + h - x}{h \cdot \left(\sqrt{x + h} + \sqrt{x}\right)}$

$= 1 + {\lim}_{h \to 0} \text{ } \frac{\cancel{h}}{\cancel{h} \cdot \left(\sqrt{x + h} + \sqrt{x}\right)}$

$= 1 + {\lim}_{h \to 0} \text{ } \frac{1}{\sqrt{x + h} + \sqrt{x}}$

At this point, you can safely plug $h = 0$ to compute the limit:

$= 1 + \frac{1}{\sqrt{x} + \sqrt{x}}$

$= 1 + \frac{1}{2 \sqrt{x}}$

Hope that this helped!