Differentiate `f(x)=x^2-5x+3` from first principle.    

Expert Answers

An illustration of the letter 'A' in a speech bubbles

To differentiate the function from first principles, we need to evaluate the limit:

`f'(x)=lim_{h->0}{f(x+h)-f(x)}/h`

Consider the numerator of the limit.

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

`=(x+h)^2-5(x+h)+3-(x^2-5x+3)`   expand brackets

`=x^2+2xh+h^2-5x-5h+3-x^2+5x-3`   collect like terms

`=2xh-5h+h^2`   factor the h

`=h(2x-5+h)`

Now put into the numerator of the limit to get:

`f'(x)=lim_{h->0}{h(2x-5+h)}/h`   cancel common factor

`=lim_{h->0}(2x-5+h)`     now...

Read
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your 48-Hour Free Trial

To differentiate the function from first principles, we need to evaluate the limit:

`f'(x)=lim_{h->0}{f(x+h)-f(x)}/h`

Consider the numerator of the limit.

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

`=(x+h)^2-5(x+h)+3-(x^2-5x+3)`   expand brackets

`=x^2+2xh+h^2-5x-5h+3-x^2+5x-3`   collect like terms

`=2xh-5h+h^2`   factor the h

`=h(2x-5+h)`

Now put into the numerator of the limit to get:

`f'(x)=lim_{h->0}{h(2x-5+h)}/h`   cancel common factor

`=lim_{h->0}(2x-5+h)`     now take the limit

`=2x-5`

The derivative of the function is `f'(x)=2x-5` .

Approved by eNotes Editorial Team