what is delta x approching? if it is approching 0 this is the definition of the derivative. so it will be 8x-3. I got this by using the rule that the derivative of ax^n=n*a*x^n-1. but they probboly want you to do it without taking the derivative.so lets start by writing it out.
y=delta*x you don't have to do this I just don't want to keep writing delta*x
(4(x+y)^2-3(x+y)-(4x^2-3x))/(y)
next we want to rewrite this so we can take the limit.
sqrare the x+y
distribute the 4
distribute the 3
get rid of some terms that cancal out and you should get 8*x+4*y-3.
now all we have to do is take the limmit as y approches 0.
i get 8x+4*0-3=8*x-3.
this verifys that the derivative of 4*x^2-3x=8*x-3
Find the following limit where f(x) = 4x^2 − 3x.
limit-->(f(x+ delta x)-f(x))/(delta x)
delta x
**delta x means the triangle symbol with x next to it***
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