Here I have a function of $2$ variables $f(x,y)$, and there is another function which is also a function of $(x,y)$ given by $k(x,y)$. The function $k$ is defined in terms of $f(x,y)$ but with functions as the parameters. SO: $$k(x,y) = f(f(xy,x), f(y,xy))$$
I am confused due to the notation, the parameters which are function $f$, which is also the outer function name. I guess it could be seen as a kind of recursion.
I wish to take the derivative of both sides, using the DEL-operator. That is the aim of this problem.
I know that there is a way via a TREE diagram to do the Chain Rule on this. But my attempt here first rests on re-labeling the function on the right side, and I want to know if this would be correct before trying to take the derivative:
Right Side: $f(f(xy,x),f(y,xy))$
$= f(f_1(xy,x), f_2(y,xy))$ where $f_1 =f(u,v)$, and $u(x,y)= xy$, $v(x,y) = x$.
Also I define $f_2 = f(s,t)$, and $s(x,y) = y$, $t(x,y) = xy$.
So now I have re-labeled this showing that there are other nested 2-variable functions.
What is the Chain Rule for applying to this kind of function?
Hope someone can help out here.
Given $z=f(u,v), u=g(x,y), v=h(x,y)$ chain rule we can write as following $$\frac{\partial}{\partial x}f(g(x,y),h(x,y)) = \frac{\partial f}{\partial u}\frac{\partial g}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial h}{\partial x} = \\ =f'_u\cdot g'_x +f'_v \cdot h'_x $$ Sometimes more easy then use notation $\frac{\partial}{\partial x}$, is to use numerical notation i.e. $f'_1$ or simply $f_1$, so above formula becomes: $$\frac{\partial}{\partial x}f(g(x,y),h(x,y))=f_1\cdot g_1 +f_2 \cdot h_1$$ Using for your case gives: $$\frac{\partial}{\partial x} k(x,y)=\frac{\partial}{\partial x} f(f(xy,x), f(y,xy)) = \\ =f_1(f(xy,x), f(y,xy))\cdot \Big (f_1(xy,x)\cdot y+f_2(xy,x)\cdot 1\Big)+\\ +f_2(f(xy,x), f(y,xy))\cdot \Big (f_1(y,xy)\cdot 0+f_2(y,xy)\cdot y\Big)$$
Addition. Now let's take your notations $f_1=f(u, v)$, where $u=u(x,y)=xy$ and $v(x,y)=x$. For second one $f_2=f(s, t)$, where $s(x,y)=y$ and $t(x,y)=xy$, let me note, that $u=t$ so, may be, it is more helpful to have one label here. Anyway, formula will be
$$\frac{\partial}{\partial x} k(x,y)=\frac{\partial}{\partial x} f(f(xy,x), f(y,xy)) = \frac{\partial}{\partial x}f(f_1,f_2)=\\ =\frac{\partial f}{\partial 1}(f_1,f_2)\cdot \frac{\partial }{\partial x}f_1(u,v) + \frac{\partial f}{\partial 2}(f_1,f_2)\cdot \frac{\partial }{\partial x}f_2(s,t)$$ where I used $\frac{\partial}{\partial i}$ for partial derivatives for $f$-s first and second arguments respectively for $i=1,2$.