Second order mixed derivative and chain rule

  • Thread starter Telemachus
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  • #1
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I want to find the second order derivative for [tex]f(x,y),x(u,v),y(u,v)[/tex], f depends on x and y, and x and y depends on u and v. I'm trying to find [tex]\frac{{\partial^2 f}}{{\partial v \partial u}}[/tex]


This is what I did:
[tex]\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}}[/tex]

Then:
[tex]\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 )[/tex]

Finally what I get:

[tex]\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}}[/tex]

Anyone knows if this is right?
 
Last edited:

Answers and Replies

  • #2
SammyS
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In the last three terms of the last line, you appear to have changed ∂y/∂u to ∂y/∂v .
 
  • #3
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Thank you SammyS, I knew something was wrong :D
 

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