# How do you find the derivative of sqrt(x-3) using the limit process?

Feb 5, 2016

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

#### Explanation:

The limit definition of the derivative is

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

With $f \left(x\right) = \sqrt{x - 3}$, you can compute the derivative as follows:

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

... expand the fraction with $\left(\sqrt{x + h - 3} + \sqrt{x - 3}\right)$...

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

... apply the formula $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$ to simplify the numerator...

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

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

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

... cancel $h$...

$= {\lim}_{h \to 0} \text{ } \frac{1}{\sqrt{x + h - 3} + \sqrt{x - 3}}$

... at this point, you can safely apply the limit, i.e. plug $h = 0$ in the denominator...

$= \frac{1}{\sqrt{x + 0 - 3} + \sqrt{x - 3}}$

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

Thus, your derivative is $\frac{1}{2 \sqrt{x - 3}}$.

Feb 5, 2016

(See below for process)
f(x)=sqrt(x-3) rArr color(green)(f'(x)=1/(2sqrt(x-3))

#### Explanation:

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

$\textcolor{w h i t e}{\text{XXX}} = {\lim}_{h \rightarrow 0} \frac{\sqrt{x + h - 3} - \sqrt{x - 3}}{h}$

Multiply numerator and denominator by conjugate of denominator:
$\textcolor{w h i t e}{\text{XXX}} = {\lim}_{h \rightarrow 0} \frac{\left(\sqrt{x + h - 3} - \sqrt{x - 3}\right)}{h} \times \frac{\left(\sqrt{x + h - 3} + \sqrt{x - 3}\right)}{\left(\sqrt{x + h - 3} + \sqrt{x - 3}\right)}$

and since $\left(a - b\right) \times \left(a + b\right) = \left({a}^{2} - {b}^{2}\right)$
$\textcolor{w h i t e}{\text{XXX}} = {\lim}_{h \rightarrow 0} \frac{\left(x + h - 3\right) - \left(x - 3\right)}{h \left(\sqrt{x + h - 3} + \sqrt{x - 3}\right)}$

color(white)("XXX")=lim_(hrarr0)cancel(h)/(cancel(h)(sqrt(x+h-3)+sqrt((x-3)))

$\textcolor{w h i t e}{\text{XXX}} = \frac{1}{\sqrt{x - 3} + \sqrt{x - 3}}$

color(white)("XXX")=1/(2sqrt(x-3)