# Partial derivative of the partial derivative of a function, with respect to the function itself

I'm trying to derive this partial derivative of a function $$f$$ with respect to $$u_x$$ and $$u_y$$ with the following form: $$f(u_x, u_y) = \left(\frac{\partial u_x}{\partial x}\right)^2 + \left(\frac{\partial u_y}{\partial y} \right)^2$$

where both $$u_x$$ and $$u_y$$ are functions of $$x$$ and $$y$$, i.e. $$u_x=u_x(x,y), u_y = u_y(x,y)$$. A more compact way to think about this is $$f$$ is a scalar function with a two-dimensional vector field as input, with $$u_x$$ and $$u_y$$ the components of that vector field. And I got to the point where

$$\frac{\partial f}{\partial u_x} = 2 \frac{\partial u_x}{\partial x} \frac{\partial}{\partial u_x} \frac{\partial u_x}{\partial x}$$

(similarly for partial w.r.t. $$u_y$$),

Then I'm stuck on this partial derivative on the right $$\frac{\partial}{\partial u_x} \frac{\partial u_x}{\partial x}$$, which seems to be the partial derivative of the partial derivative of the function, but with respect to the function itself. I don't really know how to proceed from here. Can anyone help with some pointers? Really appreciate the help.

• $\frac{\partial x}{\partial x} = 1$. If you assume sufficient continuity restrictions on $f$ and $u$, I think you can swap the order of partial derivatives. But that means that your terms are $0$. It think it is best to try it with specific functions so you can see what happens. Commented Apr 21, 2021 at 16:20

You said yourself that $$u_x$$ is some function of $$x$$ and $$y$$ thus its derivative $$\frac{\partial}{\partial x} u_x = f(x,y)$$ is a function of $$x$$ and $$y$$ as well. So in general the quantity
$$\frac{\partial}{\partial u_x}\bigg[f(x,y)\bigg] = 0\text{ .}$$