# How do you find f'(x) using the definition of a derivative f(x) =sqrt(x−3)?

Mar 28, 2016

Just take advantage of the ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

$f ' \left(x\right) = \frac{1}{2 \sqrt{x - 3}}$

#### Explanation:

$f \left(x\right) = \sqrt{x - 3}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\sqrt{x + h - 3} - \sqrt{x - 3}}{h} =$

$= {\lim}_{h \to 0} \frac{\left(\sqrt{x + h - 3} - \sqrt{x - 3}\right) \cdot \left(\sqrt{x + h - 3} + \sqrt{x - 3}\right)}{h \left(\sqrt{x + h - 3} + \sqrt{x - 3}\right)} =$

$= {\lim}_{h \to 0} \frac{{\sqrt{x + h - 3}}^{2} - {\sqrt{x - 3}}^{2}}{h \left(\sqrt{x + h - 3} + \sqrt{x - 3}\right)} =$

$= {\lim}_{h \to 0} \frac{x + h - 3 - x - 3}{h \left(\sqrt{x + h - 3} + \sqrt{x - 3}\right)} =$

$= {\lim}_{h \to 0} \frac{h}{h \left(\sqrt{x + h - 3} + \sqrt{x - 3}\right)} =$

$= {\lim}_{h \to 0} \frac{\cancel{h}}{\cancel{h} \left(\sqrt{x + h - 3} + \sqrt{x - 3}\right)} =$

$= {\lim}_{h \to 0} \frac{1}{\left(\sqrt{x + h - 3} + \sqrt{x - 3}\right)} =$

$= \frac{1}{\left(\sqrt{x + 0 - 3} + \sqrt{x - 3}\right)} = \frac{1}{\sqrt{x - 3} + \sqrt{x - 3}} =$

$= \frac{1}{2 \sqrt{x - 3}}$