Question

Does anyone know how to do second order partial derivatives?

Answer 1

yes, what exactly are you confused about?

Answer 2

h(x,y) = g(x+y)+g(x-y)
g: All Real #'s, Twice Differentiable
h: R^2 -> R, (d^2h/dx^2)=(d^2h/dy^2)

This is all that is given to me in the question. I dont know what it is asking.

Answer 3

lol it's sort of hard to help you if you don't know what the question is asking XD

Are they stating that the second partials of h are equal?
Or is that what we're trying to prove?

Answer 4

I think they are stating that.

Answer 5

Here are a few steps worked out, but im not sure if they will help

Answer 6

So they did a bunch of work to show that,\[\large\rm h_{xx}=g(x+y)_{xx}+g(x-y)_{xx}\]Or maybe we can use the lazy prime notation like you guys did,\[\large\rm h_{xx}=g''(x+y)+g''(x-y)\]So maybe we need to show that \(\large\rm h_{yy}\) will give us the same thing?
Otherwise I don't really see a question in this XD haha

Answer 7

\[\large\rm u=x+y\]\[\large\rm v=x-y\]So then,\[\large\rm h=g(u)+g(v)\]Taking our partial derivative with respect to y,\[\large\rm h_y=g'(u)\cdot u'+g'(v)\cdot v'\]\[\large\rm h_y=g'(u)\cdot (x+y)_y+g'(v)\cdot (x-y)_y\]\[\large\rm h_y=g'(u)\cdot (1)+g'(v)\cdot (-1)\]

Answer 8

So we have our first derivative in y,\[\large\rm h_y=g'(u)-g'(v)\]

Answer 9

Confused? :U

Answer 10

actually no, that makes sense

Answer 11

So ya, then our second partial just gives us another negative,\[\large\rm h_yy=g''(u)u'-g''(v)v'\]\[\large\rm h_{yy}=g''(u)(1)-g''(v)(-1)\]\[\large\rm h_{yy}=g''(u)+g''(v)\]

Answer 12

They told us that our second partials of h are equal,\[\large\rm h_{xx}=h_{yy}\]So then,\[\large\rm g(u)_{xx}+g(v)_{xx}\quad=\quad g(u)_{yy}+g(v)_{yy}\]I'm not sure if we can use the sloppy prime notation anymore since we're mixing the equations together.

Answer 13

Grr the site keeps crashing on me :(

Answer 14

And then uhhhh, I dunno :D
No question ... soooo... what happened?
You stumble into class late or something? XD

Answer 15

no thats just the way the TA wrote it out.
I emailed him asking if he could clarify what he was asking for in the question and his response was "yes thats correct" lol

Answer 16

lolol

Answer 17

thats college i guess

Answer 18

yes thats correct