consider f(x)=x^2+1 find f'(1) using the definition of derivative

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consider f(x)=x^2+1 find f'(1) using the definition of derivative

Mathematics
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The definition of a derivative is f(x+h)-f(x) all over h.
the "definition" being: f(x+h) - f(x) lim ----------- h right?
And yes there is a limit, forgot that part. lim as h approaches 0.

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yes that is right. is get confused when i see h^2 as part of the solution. is the x+h always squared?
What do you mean always squared?
(1+h)^2 + 1 - (1^2 + 1)//h h^2 +2h +1 +1 -1 -1//h h^2 +2h // h h(h+2)//h h+2 is what I get if I did it right....
i mean party of his answr is 2+2h+h^2-2/h 2h+h^2/5 h(2+h)/5 2+5=2 yes the limit is zero
as h->0; h+2 = 2 f'(x) = 2x; f'(1) = 2
that aint no 5....
the squared question is asked is ist the (1-h)^2. will i always sqaure the X+h?
i mean h. i always seem to hit 5
You will only square the x+h if your function has a 2nd degree in it somewhere.
ok.... f(x) = x^2 +1 whatever value x is ... is the value you use for x. :)
f(1+h) is the value you use and f(1) is the other...
so if its not to a power, you will not square it. just multiply normally?
....youve confused me
What you are basically doing is plugging in (x+h) into the function for x then subtracting the function itself and dividing by h. All while calculating the limit.
and x = 1 in this instance :)
yes to see if i understand if it was just +1, not x^2 +, would you not square it?
f(1+h) - f(1) // h [(1+h)^2 +1] - [1^2 +1] // h
So if you have f(x)=x+1 there is no squaring involved.
that is something i always wanted to know thank you. i will be posting a couple more questions of confusion
[h^2 +2h +1 +1] - [2] //h h^2 + 2h +2 - 2 //h h^2 + 2h // h h(h+2)//h lim{h->0} h+2 = 0+2 = 2
f'(x) = 2x f'(1) = 2(1) = 2 same answers :)

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