Question

Using the limit definition, how do you find the derivative of `f(x)=x^(1/3)`?

Answer 1

Use the fact that `a^3-b^3 = (a-b)(a^2+ab+b^2)`, so `a - b = (root3a-root3b)(root3a^2+root3(ab)+root3b^2)`

Explanation:

Therefore the conjugate of `root3(x+h) - root3x` is

`(root3(x+h)^2+root3(x+h)(root3x)+(root3x)^2)`.

The rest is analogous to finding the derivative of `x^(1/2)`.

I shall use the notation of the question with `x^(1/3)` rather than `root3(x)`

`lim_(hrarr0)((x+h)^(1/3)-x^(1/3))/h`

To save some space, let's do the algebra first, then find the limit.

`((x+h)^(1/3)-x^(1/3))/h = (((x+h)^(1/3)-x^(1/3)))/h * (((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)))/(((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3))`

`= ((x+h)-x)/(h((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)))`

`= h/(h((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3)))`

`= 1/((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3))`

So, we have

`lim_(hrarr0)((x+h)^(1/3)-x^(1/3))/h = lim_(hrarr0)1/((x+h)^(2/3)+(x+h)^(1/3)x^(1/3)+x^(2/3))`

`= 1/((x+0)^(2/3)+(x+0)^(1/3)x^(1/3)+x^(2/3))`

`= 1/(3x^(2/3))`

Bonus

It is also true that for positive integer `n`,

`a^n-b^n = (a-b)(a^(n-1)+a^(n-2)b+a^(n-3)b^2 + * * * +ab^(n-2)+b^(n-1))`.

This allow us to use the same general method for any `n^(th)` root.