Chain Rule Problem (Partial derivatives)

slr77
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Homework Statement


pic.PNG


Homework Equations

The Attempt at a Solution


I have the solution to this problem and the issue I'm having is that I don't understand this step:

pic2.PNG


Maybe I'm overlooking something simple but, for the red circled part, it seems to say that ∂/∂x(∂z/∂u) = (∂2z/∂u2)y+(∂2z/∂v∂u)(-y/x2). I realize that ∂2z/∂x∂u = ∂2z/∂u∂x but I can't see how either of them leads to the above. I almost understand it when I try to take the latter but apparently they take y and y/x^2 as constants and I don't understand why that is (assuming that is indeed what the solution is doing).
 
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slr77 said:
I realize that ∂2z/∂x∂u = ∂2z/∂u∂x

This is generally not true. The partial derivative with respect to x implicitly assumes y to be constant and the partial derivative with respect to u assumes v to be constant.

You are looking to apply the chain rule in its most basic form:
$$
\partial_x = \frac{\partial u}{\partial x} \partial_u + \frac{\partial v}{\partial x} \partial_v.
$$
What is being used is simply the very first line, but with ##z## replaced by ##\partial z/\partial u##.
 
Orodruin said:
This is generally not true. The partial derivative with respect to x implicitly assumes y to be constant and the partial derivative with respect to u assumes v to be constant.

You are looking to apply the chain rule in its most basic form:
$$
\partial_x = \frac{\partial u}{\partial x} \partial_u + \frac{\partial v}{\partial x} \partial_v.
$$
What is being used is simply the very first line, but with ##z## replaced by ##\partial z/\partial u##.

Oh of course, I remember how to do this now. ∂z/∂u is a function of (u,v) which are functions of (x,y) so I just apply the chain rule like usual.

I made some really bad mistakes here (especially applying Clairaut's theorem so incorrectly) but at least the problem now looks pretty straightforward. Thanks for the help.
 
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