Question

Using the limit definition, how do you differentiate `f(x)=x^2-4x+23`?

Answer 1

`2x-4`

Explanation:

The limit definition of the derivative is given by:

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

Putting `f(x)` into the formula we get:

`lim_(h->0)((x+h)^2-4(x+h)+23-x^2+4x-23)/h`

Now expend out the brackets to get:

`lim_(h->0) (x^2+2xh+h^2-4x-4h+23-x^2+4x-23)/h`

Gathering the like terms:

`lim_(h->0) (color(red)(cancelx^2)+2xh+h^2-color(blue)(cancel(4x))-4h+color(green)(cancel23)-color(red)(cancelx^2)+color(blue)(cancel(4x))-color(green)(cancel23))/h`

`=lim_(h->0)(2xh-4h+h^2)/h`

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

Now evaluate the limit and we are left with:

`=2x-4`