Question

help me with chain rule....


i have u = f(x ) ; x = yt^(-1/2)

find d^2u/dy^2.

Answer 1

looks like we got do some implicit stuff; or... simply regard t as independant; they come out different depending on what we are going for

Answer 2

can u show e the working...
this is using the tree diagram

Answer 3

i found partial u partial y = du/dx t^(-1/2)

Answer 4

then from what i get how to find partial^2 u/ partial y^2?

Answer 5

im not sure what is meant by 'tree' diagram; it sounds rather 'author' specific and not a general mathing term

Answer 6

what is with the "t"?

Answer 7

if there is a t in it, but you are taking the derivative of u wrt y isn't the 't' irrelevant?

Answer 8

depends on how y interacts with y :) right?

Answer 9

i am just asking, not saying. i am confused.

Answer 10

\[y = x\sqrt{t}\] right?

Answer 11

i read it as
\[u = f(\sqrt{t}y)\]

Answer 12

but you want the second derivative wrt y, not t

Answer 13

no its u = f (y / t^(1/2) )

Answer 14

yes you are right i didn't see the - sign sorry

Answer 15

@satelite : yes...scnd derivative wrt to y

Answer 16

it feels like we are trying to come into this halfway thru the problem; is there a bigger picture that this question fits with?

Answer 17

still as far as y taking the derivative wrt y
\[\frac{1}{\sqrt{t}}\] is a constant

Answer 18

the tree diagram that i was referring to is like this

http://www.youtube.com/watch?v=aZcw1kN6B8Y

Answer 19

and we don't know what u is.
\[u=f(x)\] tells us nothing. so it is just
\[u'=f'(\frac{y}{\sqrt{t}})\frac{1}{\sqrt{t}}\]

Answer 20

ok got that already

Answer 21

is there more to the problem than written? because this looks nothing like taking partials as in the video

Answer 22

im sorry i dont understand...
i am looking for a partials 2nd derivative...from the video is only 1st derivative..
so how do i continue from the answer u gave to get second partial wrt to y