Calculus Chain Rule

  • #1
laser1
134
20
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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  • #2
Please learn LaTeX and post equations using that format. Check out the handy guide at the bottom left of this post.
 
  • #3
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:
  • #4
Sorry, I don't see any difference between the problem-statement equation (top box) and your derived equation (bottom boxes):
1733853564462.png
 
  • #5
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
 
  • #6
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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  • #7
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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  • #8
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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