Function Composition Homework: Derivatives of z w.r.t x, y

Click For Summary

Homework Help Overview

The discussion revolves around the derivatives of a function z defined in terms of other variables u, v, and w, which are themselves functions of x and y. Participants are exploring the application of the chain rule to find the partial derivatives ∂z/∂x and ∂z/∂y.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the necessity of considering all variables that depend on x when calculating ∂z/∂x. There is an exploration of the expressions for the derivatives, with some participants suggesting a more comprehensive approach that includes all relevant variables.

Discussion Status

The discussion is active, with participants providing feedback on each other's expressions and approaches. Some guidance has been offered regarding the correct application of derivatives, and there is a recognition of the need to include all dependencies in the calculations.

Contextual Notes

There is an emphasis on the chain rule and the need to account for all variables that influence the derivatives, indicating a focus on understanding the relationships between the variables involved.

Ikastun
Messages
10
Reaction score
0

Homework Statement



Be z=F(u,v,w), u=f(x,y), v=e-αx, w=ln y, get the expression [itex]\partial[/itex]z/[itex]\partial[/itex]x, [itex]\partial[/itex]z/[itex]\partial[/itex]y.

Homework Equations



Chain rule.

The Attempt at a Solution



[itex]\partial[/itex]z/[itex]\partial[/itex]x=[itex]\partial[/itex]z/[itex]\partial[/itex]v*dv/dx=-α e-αx

[itex]\partial[/itex]z/[itex]\partial[/itex]y=[itex]\partial[/itex]z/[itex]\partial[/itex]w*dw/dy=1/y
 
Physics news on Phys.org
When calculating [itex]\frac{\partial z}{\partial x}[/itex],you should consider all variables that depend on x not only v!
 
Thank you. What about the following expression: ∂z/∂x=∂z/∂u*∂u/∂x+∂z/∂v*dv/dx?. Same to ∂z/∂y with w.
 
Ikastun said:
Thank you. What about the following expression: ∂z/∂x=∂z/∂u*∂u/∂x+∂z/∂v*dv/dx?.
Looks good. You even picked up on the fact that dv/dx is a regular (not partial) derivative.
Ikastun said:
Same to ∂z/∂y with w.
What did you get?
 
∂z/∂y=∂z/∂u*∂u/∂y+∂z/∂w*dw/dy.
 

Similar threads

Replies
8
Views
2K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K