Question

Find the derivative using first principles? : `x^3`

Answer 1

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

Explanation:

Using the limit definition of the derivative:

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

With `f(x)=x^3` we have:

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

And expanding using the binomial theorem (or Pascal's triangle) we get:

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