# Second order mixed derivative and chain rule

1. Aug 17, 2011

### Telemachus

I want to find the second order derivative for $$f(x,y),x(u,v),y(u,v)$$, f depends on x and y, and x and y depends on u and v. I'm trying to find $$\frac{{\partial^2 f}}{{\partial v \partial u}}$$

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

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

Finally what I get:

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

Anyone knows if this is right?

Last edited: Aug 17, 2011
2. Aug 17, 2011

### SammyS

Staff Emeritus
In the last three terms of the last line, you appear to have changed ∂y/∂u to ∂y/∂v .

3. Aug 18, 2011

### Telemachus

Thank you SammyS, I knew something was wrong :D