I have a function $f:\mathbb{R}^2 \to \mathbb{R}$. So we can write $f(x,y)$ where $x$ and $y$ are independant variables. Now we define the function $g(u,v) = f(u+v, uv)$.
Using the chain rule we have :
$$\frac{\partial g}{\partial u} = \frac{\partial f}{\partial x} + v \frac{\partial f}{\partial y}$$ $$\frac{\partial g}{\partial v} = \frac{\partial f}{\partial x} + u \frac{\partial f}{\partial y}$$
Thus we have :
$$\frac{\partial}{\partial v} \frac{\partial g}{\partial u} = \left ( \frac{\partial }{\partial x} + u \frac{\partial }{\partial y} \right) \left ( \frac{\partial f}{\partial x} + v \frac{\partial f}{\partial y} \right)$$
But here I have a problem when developing this expression. Indeed we have : $\frac{\partial v}{\partial v} = 1$ so I should expect that :
$$\left ( \frac{\partial }{\partial x} + u \frac{\partial }{\partial y} \right) v = 1$$
Yet for me we have : $\frac{\partial v}{\partial x} = 0$ since there isn't any occurence of $x$ in $v$. So I get that : $(\frac{\partial }{\partial x} + u \frac{\partial }{\partial y} )v = 0 \ne 1. $
So what is the problem here ? I think it comes from the fact that $x$ depends on $u$, but normally this shouldn't be the case right ? I mean $x$ and $y$ are defined at the beginning there are just the composant of the function $f$, and they are independant. Moreover if $x$ depends an $u$ it would mean that $f$ is a function of $u$, but it doesn't mean anything to differentiate at a function, moreover as said earlier $x$ and $y$ doesn't depend on anything.
Hence what is really going on here ?
Thank you !
You miss the term $\frac{\partial f}{\partial y}$ obtained from deriving $v\times \frac{\partial f}{\partial y}$ by $v$.
$\begin{align}\displaystyle \frac{\partial}{\partial v}\frac{\partial g}{\partial u} &=\frac{\partial}{\partial v}\left(\frac{\partial f}{\partial x} + v \frac{\partial f}{\partial y}\right) \\\\\displaystyle &=\frac{\partial}{\partial v}\left(\frac{\partial f}{\partial x}\right)+v\frac{\partial}{\partial v}\left(\frac{\partial f}{\partial y}\right)+\frac{\partial f}{\partial y} \\\\\displaystyle &=\left(\frac{\partial^2 f}{\partial x^2} + u\frac{\partial^2 f}{\partial x\partial y}\right)+v\left(\frac{\partial^2 f}{\partial x\partial y} + u\frac{\partial^2 f}{\partial y^2}\right)+\frac{\partial f}{\partial y}\end{align}$