# Partial derivative concept

I have a function $F(x,y,z)$, where $z=z(x,y)$. What does $\partial F/\partial x$ stand for? In my PDE book, from the equation $F(x,y,z,a,b)=0$ we arrive at $$\frac{\partial F}{\partial x} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} = 0$$

I assume the first term is the partial derivative of $F$ w.r.t. $x$, where the change in $z$ due to change in $x$ is neglected. The second term takes into account the change in $z$ due to change in $x$. If what I understood is indeed correct, $\partial F/\partial x$ ignores the fact that a change in $x$ changes $z$, which in turn changes $F$. Am I right so far?

If that is truly the case, then the partial derivative of $x+y+z$ w.r.t. $x$ should be $1$, even though $z$ might be a function of $x$. Which is really absurd. For, if $z=2x$, we should have the partial derivative as $3$.

• You are correct. When writing $\partial F/\partial x$ and $\partial F/\partial z$ we view $F$ as a function of five independent variables, $x$, $y$, $z$, $a$, $b$, and take the derivative to only one of them at a time. Sep 11 '17 at 17:05
• If you are satisfied with the answer feel free to accept and upvote it :) Sep 12 '17 at 14:13

Imagine you are climbing a mountain, so you can move freely in directions $x$ and $y$, but the height $z$ is forced by the form of the mountain as $z=f(x,y)$. Now you have some function $F(x,y,z)$ (e.g. temperature) that mathematically depends on the coordinates $x,y,z$, but physically only depends on $x,y$ by $F(x,y,f(x,y))$, since the value of $z$ depends on $x,y$.
As a mathematical construct, you can compute how the temperature varies with in direction west-east at sea level, i.e. without changing height. To measure that in the physical world you would need to excavate a tunnel in the mountain, but mathematically it is perfectly possible to fix $z=0$ (or any other level $z=k$) and compute $\frac{\partial F}{\partial x}$.
Now you go back to the physical, real world and realize that, in practice, you cannot change $x$ without changing $z$, so you have the formula for the total derivative that you have written.
In your example $F(x,y,z)=x+y+z$, you first see that, if you imagine you would be able to change $x$ and keep constant $z$, you have the partial derivative $\frac{\partial F}{\partial x}=1$. But then you go back to reality, notice the dependence of $z$ on $x$ as $z=2x$, and compute the total derivative $\frac{d F}{d x}=3$.