# Second order mixed derivative and chain rule

• Aug 17th 2011, 03:24 PM
Ulysses
Second order mixed derivative and chain rule
I want to find the second order derivative for $\displaystyle f(x,y),x(u,v),y(u,v)$, f depends on x and y, and x and y depends on u and v. I'm trying to find $\displaystyle \frac{{\partial^2 f}}{{\partial v \partial u}}$

This is what I did:
$\displaystyle \frac{{\partial f}}{{\partial u}}=\frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial u}}+\frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial u}}$

Then:
$\displaystyle \frac{{\partial^2 f}}{{\partial v \partial u}}=\frac{{\partial}}{{\partial v}} \left (\frac{{\partial f}}{{\partial x}}\frac{{\partial x}}{{\partial u}}\right )+\frac{{\partial}}{{\partial v}} \left (\frac{{\partial f}}{{\partial y}}\frac{{\partial y}}{{\partial u}}\right )$

Finally what I get:

$\displaystyle \displaystyle\frac{{\partial^2 f}}{{\partial v \partial u}}=\frac{{\partial^2 f}}{{\partial x^2}}\frac{{\partial x}}{{\partial v}}\frac{{\partial x}}{{\partial u}}+\frac{{\partial^2 f}}{{\partial y \partial x}} \frac{{\partial y}}{{\partial v}}\frac{{\partial x}}{{\partial u}}+\frac{{\partial f}}{{\partial x}}\frac{{\partial^2 x}}{{\partial v \partial u}}+\frac{{\partial^2 f}}{{\partial x \partial y}}\frac{{\partial x}}{{\partial v}}\frac{{\partial y}}{{\partial v}}+\frac{{\partial^2 f}}{{\partial y^2}}(\frac{{\partial y}}{{\partial v}})^2+\frac{{\partial f}}{{\partial y}}\frac{{\partial^2 y}}{{\partial v^2}}$

Anyone knows if this is right?
• Aug 17th 2011, 05:31 PM
SammyS
Re: Second order mixed derivative and chain rule
See my reply on that other site on which you posted this.