Expanding (x+h)^(1/3) using the binomial theorem

[JasonRox]

x^(1/3)

The only way I have for this on this is to show that the derivative of x^n equals Nx^(n-1).

I can't think of a way to do it in first principles.

If I can just break up (x+h)^1/3, I'm good.

If you break that up, you get...

$$x^\frac{1}{3} + 1/3x^\frac{2}{9}h^\frac{1}{9} + 1/3x^\frac{1}{9}h^ \frac{2}{9}+ h^\frac{1}{3}$$

It's probably wrong, but that is all I can think of.

Any pointers would help.

 

[faust9]

You're just trying to get the derviative of $x^{1/3}$

You have the procedure written down correctly in the second line then you went way overboard after that.

$$\frac{d}{dx}x^n=nx^{n-1}$$
where n is a constant

Replace n with 1/3--that's all there is to it.

Is your question written correctly?

 

[Fredrik]

I don't think he's trying to calculate the derivative. He's trying to prove that the rule works even when n is not an integer (specifically when n=1/3).

Jason, are you allowed to use what you know about how to take the deriviate of exponentials, logarithms and composite functions? If you are, I suggest that you use

$$x^{\frac{1}{3}}=e^{\frac{1}{3}\ln x}$$

 

[JasonRox]

Actually know we can not do that.

This must be used doing first principles. I know how to find derivatives.

We must use h approaches 0 or x approaches a.

That is why I wrote (x+h)^1/3, which is from the h approaches 0 method.

Obviously we have (x+h)^1/3-x^1/3 for the numerator. That is why I broke it up in the first post, so I can get rid of the h in the numerator, but since I can't do that because (h^1/3)/h = 1/h^2/3, according to exponent laws. This is where I am stuck.

If I had an exam right now, I would have to do the x^n -> Nx^(n-1) method, but that is not practical nor what they are asking for.

Any tips are appreciated.

Note: We must assume we don't "know" the derivative of ln x in this case.

 

[JasonRox]

So I guess I'm not the only who is stuck.

 

[arildno]

I remember a thread here at PF recently:
Set:
$$a=(x+h)^{\frac{1}{3}},b=x^{\frac{1}{3}}$$
Then, for the derivative:
$$\frac{a-b}{h}=\frac{(a-b)(a^{2}+ab+b^{2})}{(a^{2}+ab+b^{2})h}=$$
$$\frac{a^{3}-b^{3}}{(a^{2}+ab+b^{2})h}=\frac{h}{(a^{2}+ab+b^{2})h}=$$
$$\frac{1}{(a^{2}+ab+b^{2})}\to\frac{1}{3x^{\frac{2}{3}}} (h\to{0})$$

 

[JasonRox]

You're my hero!

Thanks!

Gave me a whole new perspective on how to do this.