Using the limit definition, how do you find the derivative of #f(x)= x^2 -5x + 3#?

1 Answer
Jun 11, 2018

#f'(x) = 2x - 5#

Explanation:

Given: #f(x) = x^2 - 5x + 3# use the limit definition of the derivative.

Limit definition of the derivative:

#f'(x) = lim h-> 0 " "(f(x+h) - f(x))/(h#

Use substitution to find #f(x + h):#

#f(x + h) = (x + h)^2 - 5(x + h) + 3#

#= x^2 + 2xh + h^2 -5x -5h + 3#

#f'(x) = lim h-> 0 " "(x^2 + 2xh + h^2 -5x -5h + 3 - (x^2 - 5x + 3))/h#

Distribute the negative:

#f'(x) = lim h-> 0 " "(cancel(x^2) + 2xh + h^2 cancel(-5x) -5h + cancel(3) cancel(- x^2) + cancel(5x) cancel(- 3))/h#

#f'(x) = lim h-> 0 " "(2xh + h^2 - 5h)/h#

Factor #h# from the numerator:

#f'(x) = lim h-> 0 " "(cancel(h)(2x + h - 5))/cancel(h)#

#f'(x) = lim h-> 0 " "2x + h - 5 #

Take the limit (let #h -> 0): f'(x) = 2x - 5#