# What is the derivative of sqrt(x) ?

Jun 8, 2016

$\frac{d}{\mathrm{dx}} \sqrt{x} = \frac{1}{2 \sqrt{x}}$

#### Explanation:

In general, the derivative of ${x}^{n}$ is $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$. From that, we have

$\frac{d}{\mathrm{dx}} \sqrt{x} = \frac{d}{\mathrm{dx}} {x}^{\frac{1}{2}} = \frac{1}{2} {x}^{\frac{1}{2} - 1} = \frac{1}{2} {x}^{- \frac{1}{2}} = \frac{1}{2 {x}^{\frac{1}{2}}} = \frac{1}{2 \sqrt{x}}$

We can also use the definition of the derivative:

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

Given $f \left(x\right) = \sqrt{x}$, we can find the derivative as follows:

$\frac{d}{\mathrm{dx}} \sqrt{x} = {\lim}_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$

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

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

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

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

$= \frac{1}{\sqrt{x} + \sqrt{x}}$

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