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you want the derivative of the derivative :)

how do i do that!

or the 3rd derivative..... hard to tell

implicit or solve for y

implicits.....

i dont know what d^2y/dx^2 means

d^2y/dx^2 is just the 2nd derivative of an equation, for example:
y = 2x^2
dy/dx = 4x
d2y/dx2 = 4

oh, so i am just finding the 3rd derivative pretty much?

thats what I think..... and youll have to use implicit stuff
y'=y^2(6-2x)

I forgot what implicits are):

dy = -dx -dx y//x ; dx = 1 so..
dy = -1 -y //x

y' = y^2(6-2x)
y'' = y' 2y(6-2x) + y^2(-2x)x' ; x' = 1
y'' = y' 12y -4xy - 2xy^2 is what I get

so it will be 2y y' ?

thats absolutely correct :)

after your done ; substitute the value for y' into the finished work

so if i do this problem y^2(6-x) i have to do 2y y'(6-2x)(-2)?

i find it ironic that they title me "superhero", and your named "helpless" lol

Hahaha, that is funny!

y^2(6-x) is simply the product rule with a little extra stuff jambed between the lines....

rl' + r'l is all it amounts to

r = y^2 r' = y' 2y
l = (6-x) l' = -1

oh! so it'll be 2y y'(6-2x)+y^2(-2)? or am i going off on the wrong track with the y's again?

squish it all together and simplify

sorry the problem was actually y^2(6-2x)

yes.... i missed the 2 from up top...but yes...thats it

you already know what y' equals, it was the original problem right?

I think they ask for y'' and give you y'=yadayada..... so this would be the final step :)

No y'=y^2(6-2x) so then 2y y'(6-2x)+y^2(-2) is y"

so i just leave it like that?

you can...but just be aware that any answers on a test might be written in the full expanded form.

how would i do that?

just replace the y' with y^2(6-2x) and work the math :)
2y (y^2)(6-2x)(6-2x)+y^2(-2)

2y^3(36-24x +4x^2) -2y^2

and so on and so forth...

how'd you get this part?

lets step thru this...
we found y'' right? what is it?

nevermind i think i get it because you did (6-2x)(6-2x) to get (36-24x +4x^2)

yes... :)

y' was the only variable we new the "value" of, so I just replaced it and did the math

thats right, with implicits they have to give you a way to know the x and y values.....

*f(3)=1/4

you gotta di the inegral to find F(x) ;)

so i do the anti derivative of y'?

jsut "cross multiply" to split the variables......
yeah

dy y^2(6-2x)
--- = ---------
dx 1

[S] 1/(y^2) dy = [S] 6-2x dx

whats [S] ?

its my lazy version of the elongfated "S" symbol they use for the integral/anti derivative stuff

\[\int\limits_{}\]

ohhhhhh!

i loathe the "equation" editor, slows down the computer with it trying to decode all the stuff

Hahahahah, i get you!(: how did you get that from cross multiplying though?

you seperated them?

Okay so when once i get them separated, is 1/y^2 lny^2?

now now..make it easier on yourself...1/y^2 = y^-2

[S] y^-2 dy = [S] (6-2x) dx

y^-1/-1 = 6x -(2)x^2/2 + C

-1/y = -x^2 +6x + C

why is it -2x^2/2 why is it divided by two and not just x^2?

tuen each side of..to the reciprocals.. to get "y" on top

i took typing class and passed....honest :)

what does the // mean? sorry, i dont know how to do all this math anywhere but on paper.

plug in your inital values and lets solve for "C" :)

so is it 1/4=(1/(x2-6x+c))

if the i.c is f(3)+.25

f(3)=.25

x^2 -6x +C = 4
x^2 -6x -4 = C right?

x = 3 and y = 1/4 yes

right.

9 - 18 - 4 = C C = -13 if I did it right :)

where is the 4 coming form?

lets see if: it works for us :)
1/4 = 1/(9 -18 -13)
...you put it there with the 1/4......right?

4 = 3^2 -6(3) + C

okay waitso the equation is 1/4=(1/(x^2-6x+c)?

okay so when i plug in the point i get 1/4=9-18+c?

so because 1 is on top on both sides i can cancel it?

Okay i got it now. Now i see how you got C=13 (:

.... lol good :)

okay! Thank you so much!:D

wait how do i find -C?

just change +C to -C and get an answer for it.....

wouldnt it just be -13?

so i put plus and minus 13 as my answer and my problem is completely answered?(:

+c = 13
-c = --13 = 13
c = 13

okay got it! Thank you for your help!(:

youre welcome :)

I might be back tomorrow, hahahah.(;