Question

Find the instantaneous rate of change of y with respect to x at an arbitrary value of x0.
y=2x^2

Answer 1

what is the derrivative of y = 2x^2 in respect to x?

Answer 2

have you learned that yet? Or are you doing (f(x+h) - f(x))/h ?

Answer 3

i have not covered derivatives yet

Answer 4

f(x+h) - f(x))/h

Answer 5

ok so have you done that with this yet?

Answer 6

\[\lim_{h \rightarrow 0}\frac{ 2(x+h)^2 - 2x^2 }{ h }\]

Answer 7

\[\lim_{h \rightarrow0 }\frac{ 2(x _{0}+h)^{2}-2x ^{2} }{ h }\]

Answer 8

weird...

Answer 9

ok so simplify that and wahwt do we get?

Answer 10

i'm having a hard time simplifying, my algebra is pathetic

Answer 11

ok first step, is to foil the (x+h)^2

Answer 12

2x^2+4xh+2h^2

Answer 13

?

Answer 14

\[\frac{ 2(x^2+2xh+h^2)-2x^2 }{ h }\]

Answer 15

oh ok

Answer 16

\[\frac{ 2x^2+4xh+2h^2 - 2x^2 }{ h }\]

Answer 17

ya that's where i got before, now im really stuck

Answer 18

although i know 2x^2 is cancelled

Answer 19

simplify that top and you get what?

Answer 20

4xh+2h^2

Answer 21

ok so we have (4xh+2h^2)/h

Answer 22

this is the same as (4xh)/h + (2h^2)/h right?

Answer 23

dont forget about the h on the bottom

Answer 24

ok, yes, i'm following you

Answer 25

simplify that

note: we can devide away the h, because its never 0. It just approaches 0 but never actually gets there

Answer 26

so what do we have after you devide away the h in each term?

Answer 27

4x+4?

Answer 28

not quite

Answer 29

4x+1?

Answer 30

oh nvm, just 4x

Answer 31

(4xh)/h + (2h^2)/h

lets look at the first term

4xh/h = ?

Answer 32

correct, after you run the limit.

Answer 33

do you think my poor algebra skills will hold me back much in calc1?

Answer 34

um, I hate to say it, but yeah. I dont think you have prro algebra skills, just do the things you know you can do. dont ever guess. You did all this by yourself pretty much.

Answer 35

there is lots of algebra in calc1 and calc2

Answer 36

but here in a few days or a week or so, you wont be doing this all the time. you will be able to tell me the answer to this from looking at it. what you did here way take the derivative of a function, we just went the very long way about it.


there are very easy shortcuts, but you need to understand what is going on before you can use them.

Answer 37

I got an A in precalc and an A in trig but if i had to test-into calc1 i bet i would not be admitted into the class...i feel like something is wrong with me...

Answer 38

im hard-working and i never miss class and all that stuff but i dont seem to absorb anything

Answer 39

nah, your fine. Everyone feels that way. You will now start to see why they tought you all that stuff. and its easier to remember once you know why you need to know it.

Answer 40

thanks a lot for the help...

Answer 41

np, just keep doing these problems, and it will become 2nd nature I promise.

Answer 42

will-do