Chain rule.

  • Thread starter sphyics
  • Start date
  • #1
sphyics
102
0
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
8,078
547
Assuming T and v are both functions of x, the chain rule gives:
du/dx = (∂u/∂T)(dT/dx) + (∂u/∂v)(dv/dx)
 
  • #3
sphyics
102
0
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
JHamm
391
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
sphyics
102
0
[itex]
Can you find [itex] \frac{\partial F}{\partial y} [/itex]? :)

= 2x - x/y2

i think that's the right answer :)
 
  • #6
JHamm
391
1
I think so too :)
 
  • #7
sphyics
102
0
Appreciate ur help :)
 

Suggested for: Chain rule.

  • Last Post
Replies
2
Views
133
Replies
6
Views
338
  • Last Post
Replies
6
Views
723
  • Last Post
Replies
5
Views
646
  • Last Post
Replies
2
Views
444
  • Last Post
Replies
1
Views
641
  • Last Post
Replies
2
Views
852
  • Last Post
Replies
4
Views
716
  • Last Post
Replies
2
Views
679
  • Last Post
Replies
1
Views
1K
Top