# Question #c8b78

Feb 16, 2017

$\frac{d}{\mathrm{dx}} \left(\sqrt[3]{x}\right) = {\lim}_{h \to 0} \frac{\sqrt[3]{x + h} - \sqrt[3]{x}}{h} = \frac{1}{3 \sqrt[3]{{x}^{2}}}$

#### Explanation:

Using the definition of derivative we have:

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

Now use the identity:

$\left({a}^{3} - {b}^{3}\right) = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

with $a = \sqrt[3]{x + h}$ and $b = \sqrt[3]{x}$:

$\left({\left(\sqrt[3]{x + h}\right)}^{3} - {\left(\sqrt[3]{x}\right)}^{3}\right) = \left(\sqrt[3]{x + h} - \sqrt[3]{x}\right) \left(\sqrt[3]{{\left(x + h\right)}^{2}} + \sqrt[3]{x \left(x + h\right)} + \sqrt[3]{{x}^{2}}\right)$

$\left(x + h\right) - x = \left(\sqrt[3]{x + h} - \sqrt[3]{x}\right) \left(\sqrt[3]{{\left(x + h\right)}^{2}} + \sqrt[3]{x \left(x + h\right)} + \sqrt[3]{{x}^{2}}\right)$

$h = \left(\sqrt[3]{x + h} - \sqrt[3]{x}\right) \left(\sqrt[3]{{\left(x + h\right)}^{2}} + \sqrt[3]{x \left(x + h\right)} + \sqrt[3]{{x}^{2}}\right)$

So:

$\frac{\sqrt[3]{x + h} - \sqrt[3]{x}}{h} = \frac{1}{\sqrt[3]{{\left(x + h\right)}^{2}} + \sqrt[3]{x \left(x + h\right)} + \sqrt[3]{{x}^{2}}}$

and then:

${\lim}_{h \to 0} \frac{\sqrt[3]{x + h} - \sqrt[3]{x}}{h} = {\lim}_{h \to 0} \frac{1}{\sqrt[3]{{\left(x + h\right)}^{2}} + \sqrt[3]{x \left(x + h\right)} + \sqrt[3]{{x}^{2}}} = \frac{1}{3 \sqrt[3]{{x}^{2}}}$