# How do I find derivatives of radicals like sqrt(x)?

Oct 11, 2014

We can use the Power Rule and the Difference Quotient ( First Principles ).

Power Rule

$f \left(x\right) = \sqrt{x} = {x}^{\frac{1}{2}}$

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

Difference Quotient ( First Principles )

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

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

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

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

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

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

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

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

$f ' \left(x\right) = \frac{1}{\sqrt{x + 0} + \sqrt{x}}$

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

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

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