Chain rule.

  • Thread starter sphyics
  • Start date
  • #1
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If a function is given by u = u(T,v) how to use the chain rule to write how u changes with respect to T & v.
Please specify the steps involved.
i understand chain rule as [itex]\frac{du}{dx}[/itex] = [itex]\frac{du}{dy}[/itex] [itex]\frac{dy}{dx}[/itex]
 

Answers and Replies

  • #2
mathman
Science Advisor
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Assuming T and v are both functions of x, the chain rule gives:
du/dx = (∂u/∂T)(dT/dx) + (∂u/∂v)(dv/dx)
 
  • #3
102
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Assuming T and v are both functions of x, the chain rule gives:
du/dx = (∂u/∂T)(dT/dx) + (∂u/∂v)(dv/dx)
very much thanks for ur quick response :)

how does the symbol differs in meaning with d
 
  • #4
390
1
[itex] \partial [/itex] (read "partial") is basically the equivalent of [itex] d [/itex] in multivariable calculus, it means you take the derivative of a function with respect to a variable while considering all other variables as constants during that operation.

For example if I have
[tex] F = x^2 + 2xy + \frac{x}{y} [/tex]
Then
[tex] \frac{\partial F}{\partial x} = 2x + 2y + \frac{1}{y} [/tex]
Can you find [itex] \frac{\partial F}{\partial y} [/itex]? :)
 
  • #5
102
0
[itex]
Can you find [itex] \frac{\partial F}{\partial y} [/itex]? :)
= 2x - x/y2

i think thats the right answer :)
 
  • #6
390
1
I think so too :)
 
  • #7
102
0
Appreciate ur help :)
 

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