Second Order Partial Differentiation

Question

I don't have a clue on how to start this question. I have a feeling I will need to use the Clairaut's theorem: $f_xy=f_yx$

Can anyone advise?

Answer 1

Looks like you're trying to derive the Laplacian in plane polar coordinates given the 2D cartesian form. Start with redefining $z = f(x,y) $ as $z = g(r, \theta) $ and using the chain rule.

I would recommend you begin by computing the partial derivatives $$ \frac{\partial x}{\partial r}, \frac{\partial y}{\partial r}, \frac{\partial y}{\partial \theta}, \frac{\partial x}{\partial \theta} $$

And then using the chain rule to find $$ \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} $$ and then applying it again to find the second partial derivatives and the answer should pop out.

Answer 2

$$ \partial_x z = \partial_x r\partial_r z + \partial_x \theta \partial_\theta z $$ similarly for $y$ then do it a second time

$$ \partial_{xx} z = \partial_x \left(\partial_x z\right) = \left(\partial_x r\partial_r + \partial_x \theta \partial_\theta \right)\left(\partial_x r\partial_r + \partial_x \theta \partial_\theta \right)z $$

(apologies for my lazy notation) I am also not going any further as it is mechanical from here and you should be able to do it :)