<s>[INST] Finding derivative of $\sqrt[3]{x}$ using only limits I need to finding derivative of $\sqrt[3]{x}$ using only limits
So following tip from yahoo answers: I multiplied top and bottom by conjugate of numerator
$$\lim_{h \to 0} \frac{\sqrt[3]{(x+h)} - \sqrt[3]{x}}{h} \cdot \frac{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$
$$= \lim_{h \to 0} \frac{x+h-x}{h(\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2})}$$
$$= \lim_{h \to 0} \frac{1}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$
$$= \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{x^2}}$$
$$= \frac{1}{2 \sqrt[3]{x^2}}$$
But I think it should be $\frac{1}{3 \sqrt[3]{x^2}}$ (3 instead of 2 in denominator?)
UPDATE
I found that I am using the wrong conjugate in step 1. But this (wrong) conjugate gives the same result when I multiply the numerator by it. So whats wrong with it? (I know its wrong, but why?)
 [/INST] $$\lim_{h \to 0} \frac{{(x+h)^{\frac{1}{3}}} - {x}^{\frac{1}{3}}}{h} $$
$$=\lim_{h \to 0} 

\frac{{(x+h)^{\frac{1}{3}}} - {x}^{\frac{1}{3}}}{h}
\cdot \frac{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}}{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}} $$
$$=\lim_{h \to 0} \frac{x+h-x}{h((x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3})}$$
$$=\lim_{h \to 0} \frac{1}{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}}$$
$$=\frac{1}{(x)^{2/3} + x^{1/3}(x)^{1/3} + (x)^{2/3}}$$
$$=\frac{1}{3x^{2/3}}$$
$$=\frac{x^{-2/3}}{3}$$
As obtained from the $Dx^{n} = n.x^{n-1}$
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