# How do you find f'(x) using the definition of a derivative for f(x)=sqrt(2x-1)?

Oct 5, 2015

See the explanation.

#### Explanation:

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

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

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

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

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

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

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

$= {\lim}_{h \to 0} \frac{2}{2 \sqrt{2 x - 1}} = \frac{1}{\sqrt{2 x - 1}}$