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.


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 $x=p$.

  • $\begingroup$ That's insanity. I'll do it. But it's insanity. $\endgroup$ – Addem Sep 16 '14 at 3:17
  • $\begingroup$ Well, it works equally well for the derivative of $x^{1/n}$, where $n$ is a positive integer. It is the natural generalization of the calculation of the derivative of $x^{1/2}$ that you are undoubtedly familiar with. That one uses the identity $a^2-b^2=(a-b)(a+b)$. $\endgroup$ – André Nicolas Sep 16 '14 at 3:22
  • $\begingroup$ True, it just seems like this requires knowing something very obscure--something it would be unreasonable to expect of a typical student, given the level of the material. I guess a student could expect that this would involve $a^{5}-b^{5}$ and then expect that a student would guess that this has as a factor $a-b$ and then expect that a student would use long division to find the remaining factor ... but that's some ballsy expectations. $\endgroup$ – Addem Sep 16 '14 at 3:34
  • $\begingroup$ The identity $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots +b^{n-1})$ and its relatives crop up all over the place, as sum of a finite geometric series, and in many other situations. It is not unlikely that you have already bumped into it a few times. $\endgroup$ – André Nicolas Sep 16 '14 at 3:39

$$\lim _{h\to 0}\frac{(x+h)^{1/5}-x^{1/5}}{h}$$ 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 $$\lim _{h\to 0}\frac{[(x+h)^{1/5}-x^{1/5}][(x+h)^{4/5}+(x+h)^{3/5}x^{1/5}+(x+h)^{2/5}x^{2/5}+(x+h)^{1/5}x^{3/5}+x^{4/5}]}{h[(x+h)^{4/5}+(x+h)^{3/5}x^{1/5}+(x+h)^{2/5}x^{2/5}+(x+h)^{1/5}x^{3/5}+x^{4/5}]}$$ $$=\lim _{h\to 0}\frac{[(x+h)-x]}{h[(x+h)^{4/5}+(x+h)^{3/5}x^{1/5}+(x+h)^{2/5}x^{2/5}+(x+h)^{1/5}x^{3/5}+x^{4/5}]}=\lim _{h\to 0}\frac{1}{[(x+h)^{4/5}+(x+h)^{3/5}x^{1/5}+(x+h)^{2/5}x^{2/5}+(x+h)^{1/5}x^{3/5}+x^{4/5}]}=\frac{1}{5x^{4/5}}$$


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