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

1 Answer
Mar 3, 2016

#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#