![\displaystyle \lim_{h\to 0} \frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}](/tpl/images/0270/3197/6af82.png)
![\dfrac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h} = \dfrac{\left(\sqrt[3]{x+h}-\sqrt[3]{x}\right)\left(\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}\right)}{h\left(\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}\right)}](/tpl/images/0270/3197/cb61c.png)
![\dfrac{x+h-x}{h\left(\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}\right)} = \dfrac{h}{h\left(\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}\right)} = \dfrac{1}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}}](/tpl/images/0270/3197/0cdcc.png)
![\displaystyle \lim_{h\to 0}\dfrac{1}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x(x+h)}+\sqrt[3]{x^2}} = \dfrac{1}{\sqrt[3]{(x+0)^2}+\sqrt[3]{x(x+0)}+\sqrt[3]{x^2}} = \dfrac{1}{3\sqrt{x^2}}](/tpl/images/0270/3197/c6b48.png)
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Find the derivative of the cube
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