# How do I use the limit definition of derivative to find f'(x) for f(x)=sqrt(2+6x) ?

Oct 3, 2014

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

$f \left(x\right) = \sqrt{2 + 6 x}$

f(x+h)=sqrt(2+6(x+h))=sqrt(2+6x+6h

Make the substitutions for $f \left(x\right)$ and $f \left(x + h\right)$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\sqrt{2 + 6 x + 6 h} - \sqrt{2 + 6 x}}{h}$

Rationalize the numerator

$= {\lim}_{h \to 0} \frac{\sqrt{2 + 6 x + 6 h} - \sqrt{2 + 6 x}}{h} \cdot \frac{\sqrt{2 + 6 x + 6 h} + \sqrt{2 + 6 x}}{\sqrt{2 + 6 x + 6 h} + \sqrt{2 + 6 x}}$

Remember the difference of perfect squares for the numerator

$= {\lim}_{h \to 0} \frac{\left(2 + 6 x + 6 h\right) - \left(2 + 6 x\right)}{h \cdot \sqrt{2 + 6 x + 6 h} + \sqrt{2 + 6 x}}$

Distribute the negative

$= {\lim}_{h \to 0} \frac{2 + 6 x + 6 h - 2 - 6 x}{h \cdot \sqrt{2 + 6 x + 6 h} + \sqrt{2 + 6 x}}$

Simplify numerator

$= {\lim}_{h \to 0} \frac{6 h}{h \cdot \sqrt{2 + 6 x + 6 h} + \sqrt{2 + 6 x}}$

Cancel the factors of $h$

$= {\lim}_{h \to 0} \frac{6}{\sqrt{2 + 6 x + 6 h} + \sqrt{2 + 6 x}}$

Substitute in the value of 0 for $h$ and then simplify

$= \frac{6}{\sqrt{2 + 6 x + 6 \left(0\right)} + \sqrt{2 + 6 x}}$

$= \frac{6}{\sqrt{2 + 6 x} + \sqrt{2 + 6 x}}$

$= \frac{6}{2 \sqrt{2 + 6 x}}$

$= \frac{3}{\sqrt{2 + 6 x}}$