Question

How do you use the definition of a derivative to find the derivative of `f(x)=-x^3-2x^2+3x+6`?

Answer 1

`(df)/(dx)=-3x^2-4x+3`

Explanation:

We have `f(x)=-x^3-2x^2+3x+6`

hence `f(x+h)=-(x+h)^3-2(x+h)^2+3(x+h)+6`

and therefore `f(x+h)-f(x)`

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

= `-x^3-3hx^2-3h^2x-h^3-2x^2-4hx-2h^2+3x+3h+x^3+2x^2-3x-6`

= #-cancel(x^3)-3hx^2-3h^2x-h^3-cancel(2x^2)-4hx-2h^2+cancel(3x)+3h+cancel6 +cancel(x^3)+cancel(2x^2)-cancel(3x)-cancel6#

= `-3hx^2-3h^2x-h^3-4hx-2h^2+3h`

Hence `(df)/(dx)=lim_(h->0)(f(x+h)-f(x))/h`

= `lim_(h->0)(-3hx^2-3h^2x-h^3-4hx-2h^2+3h)/h`

= `lim_(h->0)(-3x^2-3hx-h^2-4x-2h+3)`

= `-3x^2-4x+3`