Change of variables/total differential derivation

  • Thread starter thomas49th
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  • #1
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Homework Statement




http://gyazo.com/e7dc83f3f93d01391663d02bf92c0b26

Homework Equations



I am trying to derive equation 6 from 5 using 1, but I haven't got anywhere. This isn't a homework question, just an information sheet. Can someone show me the steps (no jumps please) on obtaining equation 6. After that, for practice I will attempt df/dv

Thanks
:)
 

Answers and Replies

  • #2
SammyS
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Homework Statement



attachment.php?attachmentid=45833&stc=1&d=1333420017.png

http://gyazo.com/e7dc83f3f93d01391663d02bf92c0b26

Homework Equations



I am trying to derive equation 6 from 5 using 1, but I haven't got anywhere. This isn't a homework question, just an information sheet. Can someone show me the steps (no jumps please) on obtaining equation 6. After that, for practice I will attempt df/dv

Thanks
:)
It's pretty basic.

Group the terms in (5):
First distribute [itex]\displaystyle \left.\frac{\partial f}{\partial x}\right|_y\text{ and }\left.\frac{\partial f}{\partial y}\right|_x\ .[/itex]θ

Then factor out du & dv .

[itex]\displaystyle df=\left(
\left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial u}\right|_v
+\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial u}\right|_v
\right)du +\left(
\left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial v}\right|_u
+\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial v}\right|_u
\right)dv
[/itex]



 

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  • #3
655
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It's pretty basic.

Group the terms in (5):
First distribute [itex]\displaystyle \left.\frac{\partial f}{\partial x}\right|_y\text{ and }\left.\frac{\partial f}{\partial y}\right|_x\ .[/itex]θ

Then factor out du & dv .

[itex]\displaystyle df=\left(
\left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial u}\right|_v
+\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial u}\right|_v
\right)du +\left(
\left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial v}\right|_u
+\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial v}\right|_u
\right)dv
[/itex]




I can easily get to the second equation
[itex]\displaystyle df=\left(
\left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial u}\right|_v
+\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial u}\right|_v
\right)du +\left(
\left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial v}\right|_u
+\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial v}\right|_u
\right)dv
[/itex]

but how does that become equation 6? We've gone from 4 terms to 6 and remove dv and du

Thanks
 
  • #4
SammyS
Staff Emeritus
Science Advisor
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I can easily get to the second equation
[itex]\displaystyle df=\left(
\left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial u}\right|_v
+\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial u}\right|_v
\right)du +\left(
\left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial v}\right|_u
+\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial v}\right|_u
\right)dv
[/itex]

but how does that become equation 6? We've gone from 4 terms to 6 and remove dv and du

Thanks
All that equation 6 gives is the coefficient of du.

The coefficient of dv is similar.

Compare equation 1 with [itex]\displaystyle df=\left(
\left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial u}\right|_v
+\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial u}\right|_v
\right)du +\left(
\left.\frac{\partial f}{\partial x}\right|_y \left.\frac{\partial x}{\partial v}\right|_u
+\left.\frac{\partial f}{\partial y}\right|_x \left.\frac{\partial y}{\partial v}\right|_u
\right)dv\ .[/itex]
 
  • #5
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Awesome. I see :)
 

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