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?)

Note that this works
to find the derivative
of $x^{1/n}$
where $n$
is a positive integer.
We use
$a^n-b^n
=(a-b)\sum_{k=0}^{n-1} a^k b^{n-1-k}
$.
$\begin{array}\\
\frac{(x+h)^{1/n}-x^{1/n}}{h}
&=\frac{(x+h)^{1/n}-x^{1/n}}{h}
\frac{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}\\
&=\frac{(x+h)-x}{h}
\frac{1}{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}\\
&=\frac{h}{h}
\frac{1}{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}\\
&=\frac{1}{\sum_{k=0}^{n-1} (x+h)^{k/n} x^{(n-1-k)/n}}\\
\end{array}
$
As $h \to 0$,
$\begin{array}\\
\sum_{k=0}^{n-1} (x+h)^{k/n} x^{(n-1-k)/n}
&\to \sum_{k=0}^{n-1} x^{k/n} x^{(n-1-k)/n}\\
&= \sum_{k=0}^{n-1} x^{(n-1)/n}\\
&= n x^{(n-1)/n}\\
&= n x^{1-1/n}\\
\end{array}
$ 
Therefore
$(x^{1/n})'
=\frac1{n x^{1-1/n}}
=\frac1{n} x^{1/n-1}
$.
