Partial Derivative Chain Rule

  • Thread starter jegues
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  • #1
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Homework Statement


See figure.


Homework Equations





The Attempt at a Solution



Here's what I got,

[tex]\frac{ \partial z}{\partial x} = \left( \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} \right) + \left( \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x} \right) + \left( \frac{\partial z}{\partial x}\right)[/tex]

Is this correct?
 

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Answers and Replies

  • #2
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I don't know much about partial derivatives, but it seems weird that what you're solving for (dz/dx), is also your last term on the right side of the equivalency.

EDIT: Upon further review, looks like I just don't know enough. Sorry for this not-so-helpful post :(
 
  • #3
HallsofIvy
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Yes, that is correct- although since you are given that u= g(x,y) and v= h(x), it would be better to write
[tex]\frac{\partial z}{\partial x}= \frac{\partial z}{\partial u}\frac{\partial g}{\partial x}+ \frac{\partial z}{\partial v}\frac{dh}{dx}+ \frac{\partial z}{\partial x}[/tex]
 
  • #4
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HallsofIvy,
That's exactly what I got, but it bothered me that, since [tex]\frac{\partial z}{\partial x}[/tex] appears on both sides, there doesn't seem to be any way to get an explicit value for this partial.
 

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