Derivative of x^(2/3): Help with Homework

[CalculusHelp1]

Homework Statement

Find derivative of x^(2/3) from first principles (i.e limit definition)

Homework Equations

lim h-->0 (f(x+h)-f(x)/h)

The Attempt at a Solution

[(x+h)^(2/3)-x^(2/3)]/h

I've tried multiplying the top and bottom by the conjugate, but I end up with the same equation except more to work with on the bottom and the top is a multiple of two of the original exponent (e.g after one congugate, it will be the same numerator except the power is 4/3, the next time 8/3, etc.).

I can't quite seem to figure out how to get it into a form that I can factor it or what not. Any help is appreciated.

 

[micromass]

You need to multiply by another conjugate. If you have cube roots, then you will have to use following formula

x^3-y^3=(x-y)(x^2+xy+y^2)

So try multiplying by the conjugate

(x+h)^{4/3}+x^{2/3}(x+h)^{2/3}+x^{4/3}

 

[CalculusHelp1]

Yes, that did it. Thanks a lot.