Question

How do you use the definition of a derivative to find the derivative of `f(x)=(x-6)^(2/3)`, at c=6?

Answer 1

`f'(6)` does not exist. For how to use the definition, see below. (Warning: It takes a lot of algebra to use the definition to get `f'(x)` for this function.)

Explanation:

For `f(x) = (x-6)^(2/3)`, we need some algebra to use the definition of derivative to find `f'(x)`

`f'(x) = lim_(hrarr0) (f(x+h)-f(x))/h`

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

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

Obviously (and as expected) this limit has initial form `0/0`.
We will write the difference quotient so that the numerator contains no radicals. (Often called "rationalizing" the numerator.)
Our goal is to get rid of the `h` in the denominator.

Algebra needed

The sum of two cubes can be factored:

`u^3+v^3 = (u+v)(u^2-uv+v^2)`.

(If you didn't learn this in algebra class, learn it now.)

In this question, we have `3^"rd"` roots, so we need:

`a+b = (root3a+root3b)(root3a^2-root3a root3b+root3b^2)`

`= (root3a+root3b)(root3a^2-root3(ab)+root3b^2)`

To make matters more interesting, we actually have `3^"rd"` roots of squares. So we need:

`a^2+b^2 = (root3(a^2)+root3(b^2))(root3(a^2)^2-root3(a^2) root3(b^2)+root3(b^2)^2)`

`= (root3(a^2)+root3(b^2))(root3(a^2)^2-root3(a^2b^2)+root3(b^2)^2)`

We also will use

`((x+h)-6)^2 = (x+h)^2 - 12(x+h)+36`

`= x^2+2xh+h^2-12x-12h+36`

And, of course `(x-6)^2 = x^2-12x+36`

For the derivative

We'll use the algebra discussed above to rewrite the difference quotient.
We have `a=(x+h-6)` and `b=(x-6)`.

We write:

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

`= ((root(3)((x+h-6)^2)-root(3)((x-6)^2)))/h*((root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))/((root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))`

`= ((x+h-6)^2-(x+6)^2)/(h(root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))`

`= ((x^2+2xh+h^2-12x-12h+36)-(x^2-12x+36))/(h(root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))`

`= ((x^2+2xh+h^2-12x-12h+36)-(x^2-12x+36))/(h(root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))`

`= (2xh+h^2-12h)/(h(root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))`

`= (cancel(h)(2x+h-12))/(cancel(h)(root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))`

Now we can evaluate the limit as `hrarr0` to get

`f'(x) = lim_(hrarr0)((2x+h-12))/((root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))`

`= (2x-12)/((root(3)((x-6)^2)^2 -root3((x-6)^2(x-6)^2)+root(3)((x-6)^2)^2))`

`= (2x-12)/(root(3)(x-6)^4 -root3(x-6)^4+root(3)((x-6)^4))`

`= (2(x-6))/(3root(3)(x-6)^4) = (2(x-6))/(3(x-6)root(3)(x-6)) = 2/(3root3(x-6))`

At `x=6`, the derivative `f'(x)` does not exist.