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

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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` .**