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

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SUMMARY

The forum discussion centers on the confusion surrounding an extra factor of ##\frac{1}{r} \frac{\partial z}{\partial r}## appearing in a calculus chain rule proof. The user initially presented their work for part (a) and sought assistance for part (b), where the discrepancy arose. Key insights included the importance of recognizing the partial derivative of ##z## with respect to ##x## as a function of both ##x## and ##y##, and the necessity of applying the product rule correctly when taking the second derivative with respect to ##\theta##. Ultimately, the issue was resolved by addressing these points, leading to a correct understanding of the derivation process.

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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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