Question

How do you use the definition of a derivative to find f' given `f(x)=x^3` at x=2?

Answer 1

`f'(2) = 12`

Explanation:

Use the formula `f'(x) = lim_(h-> 0) (f(x + h) - f(x))/h`.

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

Expand the binomial either by tedious algebra or by the Binomial Theorem/Pascal's Triangle.

`(x + h)^3 = (x + h)(x + h)(x + h) = (x^2 + 2xh + h^2)(x + h) = x^3 + 2x^2h + h^2x + hx^2 + 2xh^2 + h^3`

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

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

`f'(x) = lim_(h-> 0) 2x^2 + hx + x^2 + 2xh + h^2`

`f'(x) = 2x^2 + 0(x) + x^2 + 2x(0) + 0^2`

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

A check using the power rule yields similar results.

We now simply evaluate the given value within the derivative.

`f'(2) = 3(2)^2 = 12`

Hopefully this helps!