Question

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

Answer 1

The definition of a derivative is f(x+h)-f(x) all over h.

Answer 2

the "definition" being:

f(x+h) - f(x)
lim -----------
h

right?

Answer 3

And yes there is a limit, forgot that part. lim as h approaches 0.

Answer 4

yes that is right. is get confused when i see h^2 as part of the solution. is the x+h always squared?

Answer 5

What do you mean always squared?

Answer 6

(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....

Answer 7

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

Answer 8

as h->0; h+2 = 2

f'(x) = 2x; f'(1) = 2

Answer 9

that aint no 5....

Answer 10

the squared question is asked is ist the (1-h)^2. will i always sqaure the X+h?

Answer 11

i mean h. i always seem to hit 5

Answer 12

You will only square the x+h if your function has a 2nd degree in it somewhere.

Answer 13

ok.... f(x) = x^2 +1 whatever value x is ... is the value you use for x. :)

Answer 14

f(1+h) is the value you use and f(1) is the other...

Answer 15

so if its not to a power, you will not square it. just multiply normally?

Answer 16

....youve confused me

Answer 17

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.

Answer 18

and x = 1 in this instance :)

Answer 19

yes to see if i understand if it was just +1, not x^2 +, would you not square it?

Answer 20

f(1+h) - f(1) // h

[(1+h)^2 +1] - [1^2 +1] // h

Answer 21

So if you have f(x)=x+1 there is no squaring involved.

Answer 22

that is something i always wanted to know thank you. i will be posting a couple more questions of confusion

Answer 23

[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

Answer 24

f'(x) = 2x

f'(1) = 2(1) = 2 same answers :)