Question

let f(x)=2x^2-3x +1

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

Answer 1

Replace x with (x+h)
-> f(x+h) = 2(x+h)^2 - 3(x+h) + 1
Expand
-> f(x+h) = 2x^2 + 4xh + 2h^2 - 3x - 3h +1
subtract original function
->(2x^2+4xh+2h^2-3x-3h+1) - (2x^2-3x+1)
add like terms **Notice how some terms will cancel
->4xh + 2h^2-3h
Factor out an h
-> h(4x + 2h - 3)
This is important because now we can cancel the h in the denominator
Now re-evaluate the lim of (4x + 2h - 3) as h->0
lim = 4x -3

Hope this helps