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?)
 A: Here is a hint: Use the identity $(a^3-b^3)=(a-b)\cdot(a^2+ab+b^2)$ with $a$, $b$ being suitable cube roots. Otherwise, the method is similar to the one you tried.
A: $$\lim_{h \to 0} \frac{{(x+h)^{\frac{1}{3}}} - {x}^{\frac{1}{3}}}{h} $$
$$=\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 $Dx^{n} = n.x^{n-1}$
A: I understand that the point of this exercise is to apply the limit definition of the derivative to a function where the limit calculation is "tricky". But it's worth noting that if $F(x,y)=0$ identically (as in $y-\sqrt[3]{x}=0$ in this problem) then $\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$. 
So given that $x=y^3$, we have that $\frac{dx}{dy}=3y^2$ (either using the power rule or a simpler limit computation). That makes $\frac{dy}{dx}=\frac{1}{3y^2}=\frac{1}{3(\sqrt[3]{x})^2}=\frac{1}{3}x^{-\frac{2}{3}}$.
A: You can use a similar 'trick' to find the derivative of $ y=\sqrt[n] x $. The limit will involve multiplying the numerator $(x+h)^{1/n} - x^{1/n} $ by an appropriate expression to get $ (x + h) - x $ .
A: 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}
$.
