Find the derivative of x^1/5 from the definition I've been trying to figure out how to compute the derivative of $f(x) = x^{1/5}$ at $x=1$ from the definition.  Here's what I've done:
$$f'(1) = \displaystyle \lim_{\Delta x\rightarrow 0} \frac{f(1+\Delta x)-f(1)}{\Delta x}$$
$$=\displaystyle \lim_{\Delta x\rightarrow 0} \frac{(1+\Delta x)^{1/5}-1}{\Delta x}$$
At this point, I thought about multiplying by the conjugate $(1+\Delta x)^{1/5}-1$ but that doesn't get very far, and nor does multiplying by the conjugate of the result.  
I then thought of multiplying by $(1+\Delta x)^{4/5}$ and got 
$$\displaystyle \lim_{\Delta x\rightarrow 0}\frac{(1+\Delta x)^{1/5}-1}{\Delta x}\frac{(1+\Delta x)^{4/5}}{(1+\Delta x)^{4/5}}=\lim_{\Delta x\rightarrow 0}\frac{1+\Delta x-(1+\Delta x)^{4/5}}{\Delta x}$$
$$\displaystyle = 1+\lim_{\Delta x\rightarrow 0}\frac{1-(1+\Delta x)^{4/5}}{\Delta x}$$
and that doesn't look like progress to me.  I thought about expressing $\Delta x = \Delta x^{5/5}$ and distributing into the fraction, from the start of the problem.
$$\displaystyle \lim_{\Delta x \rightarrow 0}(\Delta x^{-5}+\Delta x^{-4})^{1/5} - \frac{1}{\Delta x}$$
and again I don't seem to be making progress.  
I tried searching for this problem on Google and in this Stackexchange and didn't see anything quite like it.  Help would be appreciated.
 A: Use the fact that $a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$.
Take $a=(1+\Delta x)^{1/5}$ and $b=1$. 
So we need to multiply top and bottom by $a^4+a^3b+a^2b^2+ab^3+b^4$. 
On top we end up with $a^5-b^5$, which is $\Delta x$, and cancels the $\Delta x$ at the bottom. 
We end up needing to find the limit as $\Delta x$ approaches $0$ of 
$$\frac{1}{a^4+a^3b+a^2b^2+ab^3+b^4}.$$
Note that $a\to 1$, and $b=1$, so at the bottom we get $5$ terms all with limit $1$.
The same basic strategy will give us the derivative of $x^{1/5}$ at $x=p$. 
A: $$\lim _{h\to 0}\frac{(x+h)^{1/5}-x^{1/5}}{h}$$
Multiply and dived it by $(x+h)^{4/5}+(x+h)^{3/5}x^{1/5}+(x+h)^{2/5}x^{2/5}+(x+h)^{1/5}x^{3/5}+x^{4/5}$ then
$$\lim _{h\to 0}\frac{[(x+h)^{1/5}-x^{1/5}][(x+h)^{4/5}+(x+h)^{3/5}x^{1/5}+(x+h)^{2/5}x^{2/5}+(x+h)^{1/5}x^{3/5}+x^{4/5}]}{h[(x+h)^{4/5}+(x+h)^{3/5}x^{1/5}+(x+h)^{2/5}x^{2/5}+(x+h)^{1/5}x^{3/5}+x^{4/5}]}$$
$$=\lim _{h\to 0}\frac{[(x+h)-x]}{h[(x+h)^{4/5}+(x+h)^{3/5}x^{1/5}+(x+h)^{2/5}x^{2/5}+(x+h)^{1/5}x^{3/5}+x^{4/5}]}=\lim _{h\to 0}\frac{1}{[(x+h)^{4/5}+(x+h)^{3/5}x^{1/5}+(x+h)^{2/5}x^{2/5}+(x+h)^{1/5}x^{3/5}+x^{4/5}]}=\frac{1}{5x^{4/5}}$$
