Question

How do you find f'(x) using the definition of a derivative for `f(x)=(x-6)^(2/3)`?

Answer 1

Short answer: use `a^3-b^3 = (a-b)(a^2+ab+b^2)` to "rationalize" the numerator.

Explanation:

We know that `a^2-b^2 = (a-b)(a+b)` and we use it to "rationalize" a difference of square roots.
As in: `2/(3-sqrt5) = (2(3+sqrt5))/(9-5)`

For the crucial step in this problem, we use the fact that
`a^3-b^3 = (a-b)(a^2+ab+b^2)`.
This allows us to "rationalize" a difference of cube roots.

`a-b = (root(3)a - root(3)b)(root(3)a ^2+root(3)a root(3)b+root(3)b^2)`

In this case we have cube roots of squares, but the idea is the same.

From the identity above, we get:

`(a^(2/3)-b^(2/3)) (a^(4/3)+(ab)^(2/3)+b^(4/3)) = a^2-b^2`

`((x+h-6)^(2/3)-(x-6)^(2/3)) ((x+h-6)^(4/3)+((x+h-6)(x-6))^(2/3)+(x-6)^(4/3))`

`= (x+h-6)^2-(x-6)^2`

`= ((x-6)+h)^2 - (x-6)^2`

`= (x-6)^2+2(x-6)h +h^2 - (x-6)^2`

`= 2h(x-6)`

Using the above algebra, we can find:

`lim_(hrarr0)((x+h-6)^(2/3)-(x-6)^(2/3))/h`

`= lim_(hrarr0)(2h(x-6))/(h((x+h-6)^(4/3)+((x+h-6)(x-6))^(2/3)+(x-6)^(4/3)))`

`= (2(x-6))/(3(x-6)^(4/3)) = 2/(3(x-6)^(1/3))`