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

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The discussion revolves around a calculus problem involving the chain rule, where a user encounters an unexpected extra factor of ##\frac{1}{r} \frac{\partial z}{\partial r}## in their proof. They seek assistance in identifying the flaw in their derivation. Other participants point out that the partial derivative of ##z## with respect to ##x## depends on both ##x## and ##y##, and emphasize the importance of considering cross terms and the product rule when calculating second derivatives. After some clarification and corrections, the user acknowledges that the issue has been resolved. The conversation highlights the need for careful attention to derivative rules in calculus proofs.
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Homework Statement
See description
Relevant Equations
Chain Rule
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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 :(

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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
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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.
 
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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