Is \frac{∂y}{∂x}×\frac{∂x}{∂z}=-\frac{∂y}{∂z}?

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Discussion Overview

The discussion centers around the relationship between partial derivatives and the chain rule, specifically questioning whether the equation \(\frac{∂y}{∂x} \times \frac{∂x}{∂z} = -\frac{∂y}{∂z}\) holds true. Participants explore this concept through examples and conditions under which certain relationships may or may not apply.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that the equation involving a minus sign is not valid according to the chain rule.
  • Others propose that under specific conditions, such as when the variables are related by a function \(G(x,y,z)=0\), certain relationships can be derived, leading to the equation \(\frac{\partial{Y}}{\partial{x}} \frac{\partial{X}}{\partial{z}} \frac{\partial{Z}}{\partial{y}} = -1\).
  • A participant provides an example using the function \(G(x,y,z)=x+y+z\) to illustrate how the condition \(G=0\) can be satisfied identically, leading to specific values for the partial derivatives.
  • Another participant questions whether there is a general formula for partial derivatives or if it varies based on different conditions.
  • Concerns are raised about the clarity of statements made regarding the derivation of \(z\) in terms of \(x\), \(y\), and \(w\), with one participant deeming it meaningless.

Areas of Agreement / Disagreement

Participants generally disagree on the validity of the original equation and its relation to the chain rule. Multiple competing views remain regarding the conditions under which certain relationships may hold.

Contextual Notes

Participants express uncertainty about the general applicability of the discussed relationships and whether they can be derived from the chain rule under all circumstances.

tade
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Is [tex]\frac{∂y}{∂x}×\frac{∂x}{∂z}=-\frac{∂y}{∂z}[/tex]?
 
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No, the chain rule does not involve such a minus sign.
Why are you asking?
 
It is true that IF the 3 variables x,y,z are related together in a sufficiently nice condition G(x,y,z)=0, THEN, we may solve for, in a region about a solution point, one of the variables in terms of the other two, i.e, functions X(y,z), Y(x,z), Z(x,y) exist, so that within that region we have that G(X(y,z),y,z)=G(x,Y(x,z),z)=G(x,y,Z(x,y)=0 identically.

In these cases, it is true that we have the counter-intuitive result:
[tex]\frac{\partial{Y}}{\partial{x}}\frac{\partial{X}}{\partial{z}}\frac{\partial{Z}}{\partial{y}}=-1[/tex]
 
To give a simple example.
Consider the function G(x,y,z)=x+y+z

Now, the condition G(x,y,z)=0 gives rise to the equation x+y+z=0
We may now form three separate function definitions:
X(y,z)=-y-z
Y(x,z)=-x-z
Z(x,y)=-x-y

We have now that:
G(X(y,z),y,z)=-y-z+y+z=0, i.e, the condition G=0 is satisfied IDENTICALLY, for all choices of y and z.
Similarly with the other two substitutions.

We see that in this case, that we have:
[tex]\frac{\partial{Y}}{\partial{x}}=\frac{\partial{X}}{\partial{z}}=\frac{\partial{Z}}{\partial{y}}=-1[/tex]
and therefore,
[tex]\frac{\partial{Y}}{\partial{x}}\frac{\partial{X}}{\partial{z}}\frac{\partial{Z}}{\partial{y}}-1[/tex]
 
CompuChip said:
No, the chain rule does not involve such a minus sign.
Why are you asking?
Try using the simple example z = x + y

Isn't there a minus sign?
 
In that specific case, the equation is true but it is NOT "the chain rule". Your initial post implied that you were offering this as a general formula derived from the chain rule.
 
HallsofIvy said:
In that specific case, the equation is true but it is NOT "the chain rule". Your initial post implied that you were offering this as a general formula derived from the chain rule.
Is there a general formula for partial derivatives or is it a collection of several formulas based on different conditions?

Anyway, consider a function w ( x, y, z ). We can then derive z ( x, y, w ).

In this case, is
[tex]\frac{∂w}{∂z}×\frac{∂z}{∂x}=-\frac{∂w}{∂x}?[/tex]
 
"w ( x, y, z ). We can then derive z ( x, y, w )."
This is meaningless.
 

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