Chain Rule Application: Solving a First Year Uni Math Problem

Click For Summary

Homework Help Overview

The discussion revolves around applying the chain rule in the context of a first-year university mathematics problem involving partial derivatives. The original poster seeks assistance in demonstrating a specific relationship between derivatives of a function w=f(u,v) where u and v are defined in terms of x and y.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the chain rule and express uncertainty about how to differentiate the variables u and v with respect to x and y. There are attempts to clarify the differentiation process and the relationships between the variables.

Discussion Status

Some participants have provided expressions for the partial derivatives of w with respect to x and y, indicating a progression in understanding. However, there remains some uncertainty regarding the differentiation of u and v, and no explicit consensus has been reached on the overall approach to the problem.

Contextual Notes

The original poster's inquiry suggests a need for clarification on the application of the chain rule, particularly in the context of the specific functions defined for u and v. There is an indication of confusion about which variables to differentiate with respect to.

3pear
Messages
4
Reaction score
0
first year uni math problem~~

  if w=f(u,v)has continuous partial derivatives and u=x+y
  and v=x-y,use the chain rule to show that
  
  (dw/dx)/(dw/dy)=(df/du)^2-(df/dv)^2


this is a first year uni math problem,is there anyone can help we with it??
thx a lot! :cry:
 
Physics news on Phys.org


do you know the chain rule?
 


yeah i know,but no idea homework do i apply on this question~~
 


[tex]\frac{\partial w}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x}[/tex]

[tex]\frac{\partial w}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y}[/tex]
 


ok,i got it,but i still not clear which varible i take respect to when i differenciate for u=x+y and v=x-y...
 


[tex]\frac{\partial w}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} = \frac{\partial f}{\partial u}*1 + \frac{\partial f}{\partial v}*1 = \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}[/tex]
 


oh!i got it~~~ thanks mates~
 

Similar threads

Replies
19
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
21
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
1K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K