Multi-variable chain rule (second derivative)

[MaryLaine]

Hi, I've been stuck with this problem for a while:

if z=z(x,y), x=x(u,v) and y=y(u,v), prove that this equation is true:

\(\displaystyle \dfrac{\partial^2 z}{\partial u^2}\, =\, \dfrac{\partial^2 z}{\partial x^2}\, \left(\dfrac{\partial x}{\partial u}\right)^2\, +\, 2\, \dfrac{\partial^2 z}{\partial x \partial y}\, \dfrac{\partial x}{\partial u}\, \dfrac{\partial y}{\partial u}\, +\, \dfrac{\partial^2 z}{\partial y^2}\, \left(\dfrac{\partial y}{\partial u}\right)^2\, +\, \dfrac{\partial z}{\partial x}\, \dfrac{\partial^2 x}{\partial u^2}\, +\, \dfrac{\partial z}{\partial y}\, \dfrac{\partial^2 y}{\partial u^2}\)

Now, I understand that this is probably just a formula, and I get a result partially, save the part that begins with a 2 (the mixed partial derivative). Could anyone explain where it comes from?

Thank you.

 

[Deleted member 4993]

Hi, I've been stuck with this problem for a while:

if z=z(x,y), x=x(u,v) and y=y(u,v), prove that this equation is true:

\(\displaystyle \dfrac{\partial^2 z}{\partial u^2}\, =\, \dfrac{\partial^2 z}{\partial x^2}\, \left(\dfrac{\partial x}{\partial u}\right)^2\, +\, 2\, \dfrac{\partial^2 z}{\partial x \partial y}\, \dfrac{\partial x}{\partial u}\, \dfrac{\partial y}{\partial u}\, +\, \dfrac{\partial^2 z}{\partial y^2}\, \left(\dfrac{\partial y}{\partial u}\right)^2\, +\, \dfrac{\partial z}{\partial x}\, \dfrac{\partial^2 x}{\partial u^2}\, +\, \dfrac{\partial z}{\partial y}\, \dfrac{\partial^2 y}{\partial u^2}\)

Now, I understand that this is probably just a formula, and I get a result partially, save the part that begins with a 2 (the mixed partial derivative). Could anyone explain where it comes from?

Thank you.

Start with:

\(\displaystyle \displaystyle{\dfrac{\partial z}{\partial u} = \dfrac{\partial z}{\partial x} \cdot \dfrac{\partial x}{\partial u} \ + \ \dfrac{\partial z}{\partial y} \cdot \dfrac{\partial y}{\partial u}}\)

 

[MaryLaine]

Start with:

\(\displaystyle \displaystyle{\dfrac{\partial z}{\partial u} = \dfrac{\partial z}{\partial x} \cdot \dfrac{\partial x}{\partial u} \ + \ \dfrac{\partial z}{\partial y} \cdot \dfrac{\partial y}{\partial u}}\)

Yeah, I already got there.

I know that twice the mixed partial derivative has something to do so that the function can be continuous, but I really don't know how to prove the formula! I just know that the formula works similarly to the binomial one... or something. If it even is a formula at all.
I've been stuck on this for hours now.