# First Principles Example 3: square root of x

## Key Questions

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

#### Explanation:

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

$f \left(x\right) = \sqrt{x + 3} , f \left(x + h\right) = \sqrt{x + h + 3}$, then

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

If we evaluate this right away, we get

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

so we need to simplify as this is an indeterminate form.

Multiply the entire limit by the numerator's conjugate, which is $\frac{\sqrt{x + h + 3} + \sqrt{x + 3}}{\sqrt{x + h + 3} + \sqrt{x + 3}}$. This is the same as multiplying by $1.$

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

The numerator becomes

$\sqrt{x + h + 3} - \sqrt{x + 3} \cdot \left[\sqrt{x + h + 3} + \sqrt{x + 3}\right] = x + h + 3 - \left(x + 3\right) = x + h + 3 - x - 3 = h$

f'(x)=lim_(h->0)(cancelx+h+cancel3-cancelx-cancel3)/(h(sqrt(x+h+3)+sqrt(x+3))

f'(x)=lim_(h->0)(cancelh)/(cancelh(sqrt(x+h+3)+sqrt(x+3))

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

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

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

• Definition

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

By Definition,

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

by multiplying the numerator and the denominator by $\sqrt{x + h} + \sqrt{x}$,

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

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

by cancelling out $x$'s and $h$'s,

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

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

I hope that this was helpful.