Question

How do I use the limit definition of derivative to find `f'(x)` for `f(x)=x^3-2` ?

Answer 1

`f'(x)=3x^2`

Explanation:

`f(x)=x^3-2`

The limit definition states that:

`f'(x)=lim_(h->0) (f(x+h)-f(x))/h`

In our case, this is:

`f'(x)=lim_(h->0)(((x+h)^3-2)-(x^3-2))/h`

Straight away, the `-2`s cancel with each other, we also expand the term in the brackets to get:

`=lim_(h->0)(x^3+3x^2h+3xh^2-x^3)/h`

The `x^3` cancel each other leaving us with:

`=lim_(h->0)(3x^2h+3xh^2)/h`

Cancel the `h`s on the top with the bottom to get:

`=lim_(h->0)(3x^2+3xh)`

Evaluating this limit, there is no `h` on the first term so it will stay put. As `h` on the 2nd term tends to 0, the second term will vanish completely to leave us with:

`f'(x)=3x^2`