Chain Rule Problem (Partial derivatives)

[slr77]

Homework Statement

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:

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).

 

[Orodruin]

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 $\partial z/\partial u$.

 

[slr77]

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.