Why Does an Extra Factor Appear in My Calculus Chain Rule Proof?

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    Calculus Chain
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Homework Help Overview

The discussion revolves around a calculus problem related to the chain rule, specifically addressing an extra factor that appears in the proof of a derived equation. The original poster expresses confusion regarding the presence of this factor in their calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the original poster's steps in the proof, questioning the derivation of the extra factor and exploring the implications of partial derivatives. There are suggestions to check specific examples to clarify the reasoning.

Discussion Status

Participants are actively engaging with the original poster's work, offering insights into potential errors and encouraging the exploration of specific cases. Some guidance has been provided regarding the need to consider additional terms from the product rule and the nature of partial derivatives.

Contextual Notes

There are mentions of formatting issues with the equations presented, as well as a focus on ensuring clarity in the mathematical expressions used in the discussion.

laser1
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Homework Statement
See description
Relevant Equations
Chain Rule
1733850880402.png


Part (a) is fine. I have attached my work for this here anyway because I use the first two lines of (a) in my proof of (b). In part (b), I get the extra factor of ##\frac{1}{r} \frac{\partial z}{\partial r}## and I don't know why. Can anyone help? I am unable to figure out the flaw in my steps :(

1733850892280.png


1733850919966.png
 
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Please learn LaTeX and post equations using that format. Check out the handy guide at the bottom left of this post.
 
renormalize said:
Please learn LaTeX and post equations using that format. Check out the handy guide at the bottom left of this post.
ok
1733852124250.png

1733854908741.png
 
Last edited:
Sorry, I don't see any difference between the problem-statement equation (top box) and your derived equation (bottom boxes):
1733853564462.png
 
renormalize said:
Sorry, I don't see any difference between the problem-statement equation (top box) and your derived equation (bottom boxes):
View attachment 354357
my bad I have edited #3
 
laser1 said:
my bad I have edited #3
The partial derivative of ##z## wrt ##x## is a function of both ##x## and ##y##. The second order derivative of ##z## wrt ##r## should have terms like ##\frac{\partial^2 z}{\partial x \partial y}##.

Try ##f(x,y) = xy## as an example to check your working.
 
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That said, the cross terms cancel. But, you are also missing the terms arising from the product rule when taking the second derivative wrt ##\theta##.
 
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PeroK said:
That said, the cross terms cancel. But, you are also missing the terms arising from the product rule when taking the second derivative wrt ##\theta##.
Good catch, it all works out now, thanks!
 
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