Partial Derivative of a function of x, y, and a function of x and y Let $F$ be a function of $x$, $y$, and $f(x,y)$.
Does $\dfrac{\partial{F}}{\partial{x}}$ take into account the change in $f$ due to the change in $x$?
For instance let $F(x,y,f(x,y)) = x+y+f(x,y)$. Then is $\dfrac{\partial{F}}{\partial{x}} = 1$ or $1+\dfrac{\partial{f}}{\partial{x}}$?
 A: This is lousy notation, and completely ambigious as it is formulated.
To begin with, saying “F is a function of $x$, $y$ and $f(x,y)$” is bound to lead to confusion.
In order to even talk about partial derivatives unambiguously, you need to specify a coordinate system $(x,y,z)$, and define your function $F$ as a function of $x$, $y$ and $z$, where these are independent variables. Then “partial derivative with respect to $x$” means that you vary $x$ but keep the other variables $y$ and $z$ constant.
So, for example, if $F(x,y,z)=x+y+z$, then $\frac{\partial F}{\partial x}(x,y,z)=1$. And thus also $\frac{\partial F}{\partial x}(x,y,f(x,y))=1$, since we substitute $z=f(x,y)$ into the function $\frac{\partial F}{\partial x}(x,y,z)$, i.e., we let $z=f(x,y)$ after we have already differentiated.
On the other hand, if you let $z=f(x,y)$ before differentiating, then you are in fact considering another function, namely the composite function
$$
G(x,y) = F(x,y,f(x,y))
.
$$
The partials of $G$ are given by the chain rule, for example
$$
\frac{\partial G}{\partial x}(x,y)
=
\frac{\partial F}{\partial x}(x,y,f(x,y))
+
\frac{\partial F}{\partial z}(x,y,f(x,y))
\,
\frac{\partial f}{\partial x}(x,y)
.
$$
This expression is sometimes also written as
$$
\frac{\partial}{\partial x}
\biggl(
F(x,y,f(x,y))
\biggr)
,
\tag{$*$}
$$
but note that this is different from
$
\frac{\partial F}{\partial x}(x,y,f(x,y))
$
that we had above. So you can't write just $\frac{\partial F}{\partial x}$ if it's ($*$) that you want to refer to.
A: For a function $F(x,y,f(x,y))=x+y+f(x,y)$, the derivative with respect to $x$ is
$$\frac{\partial}{\partial x}(x+y+f(x,y))=\frac{\partial }{\partial x}x +\frac{\partial}{\partial x}y +\frac{\partial}{\partial x} f(x,y)$$
Assuming $y$ is not a function of $x$, this means
$$1+\frac{\partial}{\partial x} f(x,y)$$
