Question

How do you use the limit definition of the derivative to find the derivative of `f(x)=-4x^3-3x^2+1`?

Answer 1

`f'(x) = -12x^2 -6x`

Explanation:

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

So for `f(x) = -4x^3 - 3x^2 + 1` we have:

`f'(x) = lim_(h->0)({ -4(x+h)^3 - 3(x+h)^2 + 1 } - { -4x^3 - 3x^2 + 1 })/h`
`:. f'(x) = lim_(h->0)({ -4 (x^3 + 3hx^2 + 3h^2x + h^3) - 3(x^2 + 2hx + h^2) + 1 } - { -4x^3 - 3x^2 + 1 })/h`

`:. f'(x) = lim_(h->0)( -4x^3 -12hx^2 -12h^2x -4 h^3 - 3x^2 -6hx -3h^2 + 1 +4x^3 + 3x^2 - 1 )/h`

`:. f'(x) = lim_(h->0)( -12hx^2 -12h^2x -4 h^3 -6hx -3h^2 )/h`

`:. f'(x) = lim_(h->0)( -12x^2 -12hx -4 h^2 -6x -3h )`
`:. f'(x) = -12x^2 -6x`