Question

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

Answer 1

Please see the explanation section below.

Explanation:

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)`

Therefore the conjugate of `a^(2/3)-b^(2/3)` is

`a^(4/3)+a^(2/3)b^(2/3)+b^(4/3)`.

And the product is `a^2-b^2`.

`lim_(hrarr0)((x+6+h)^(1/3)-(x+6)^(1/3))/h`

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

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

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

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

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

So, we have

`lim_(hrarr0)((x+6+h)^(1/3)-(x+6)^(1/3))/h= lim_(hrarr0)(2(x+6)+h)/((x+6+h)^(4/3)+(x+6+h)^(2/3)(x+6)^(2/3)+(x+6)^(4/3))`

`= (2(x+6))/((x+6)^(4/3)+(x+6)^(2/3)(x+6)^(2/3)+(x+6)^(4/3))`

`= (2(x+6))/(3(x+6)^(4/3)`

`= 2/(3(x+6)^(1/3))`

Answer 2

Use `a^3 - b^3 = (a - b)(a^2 + ab + b^2)` as a pattern to clear the `1/3` power from the numerator. Please see the explanation.

Explanation:

Use: `f'(x) = lim_(hto0)(f(x+h)-f(x))/h`

Given: `f(x) = (x - 6)^(2/3)`, then `f(x+h) = (x + h-6)^(2/3)`

This notation can get hard to follow so I am going to make the following substitution:

Let `a = f(x + h)`
Let `b = f(x)`

`f'(x) = lim_(hto0)(a-b)/h`

We want the numerator to become `a^3 - b^3`, because that will change the power of each of the terms from `2/3` to 2. Therefore, we multiply by `(a^2 + ab + b^2)/(a^2 + ab + b^2)`

`f'(x) = lim_(hto0)(a-b)/h(a^2 + ab + b^2)/(a^2 + ab + b^2)`

The numerator becomes what we want:

`f'(x) = lim_(hto0)(a^3-b^3)/(h(a^2 + ab + b^2))" [1]"`

Now we work with the numerator so that we can make a factor of h cancel out the h in the denominator.

`a^3 = (f(x + h))^3`
`a^3 = ((x + h - 6)^(2/3))^3`
`a^3 = (x + h - 6)^2`
`a^3 = x^2 +hx -6x + hx + h^2 - 6h -6x -6h - 36" [2]"`

`b^3 = (f(x))^3`
`b^3 = ((x - 6)^(2/3))^3`
`b^3 = (x - 6)^2`
`b^3 = x^2 - 12x - 36" [3]"`

Subtract equation [3] from equation [2]:

`a^3 - b^3 = x^2 +hx -6x + hx + h^2 - 6h -6x -6h - 36 - x^2 + 12x + 36`

`a^3 - b^3 = cancel(x^2 +)hx cancel(-6x) + hx + h^2 - 6h cancel(-6x) -6h cancel(- 36)cancel( - x^2) cancel(+ 12x)cancel( + 36)`

`a^3 - b^3 = 2hx + h^2 - 12h`

`a^3 - b^3 = h(2x + h - 12)" [4]"`

Substitute the right side of equation [4] into the numerator of equation [1]:

`f'(x) = lim_(hto0)(h(2x + h - 12))/(h(a^2 + ab + b^2))" [5]"`

The h in the numerator cancels the h in the denominator:

`f'(x) = lim_(hto0)(cancel(h)(2x + h - 12))/(cancel(h)(a^2 + ab + b^2))`

We can let `hto 0`. This makes "a" become "b" so the denominator becomes `3b^2`

`f'(x) = (2x - 12)/(3b^2)`

Substitute f(x) for b:

`f'(x) = (2x - 12)/(3(f(x))^2)`

Remove a factor of 2 from the numerator:

`f'(x) = 2(x - 6)/(3(f(x))^2)`

Remove a factor of 3 from the denominator:

`f'(x) = 2/3(x - 6)/(f(x))^2`

Substitute `(x - 6)^(2/3)` for `f(x)`

`f'(x) = 2/3(x - 6)/(((x - 6)^(2/3))^2)`

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

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

This is what we expected.