TITLE: The notation for partial derivatives
QUESTION [17 upvotes]: Today, in my lesson, I was introduced to partial derivatives. One of the things that confuses me is the notation. I hope that I am wrong and hope the community can contribute to my learning. In single-variable calculus, we know that, given a function $y =f(x)$, the derivative of $y$ is denoted as $\frac {dy}{dx}$. I understand this as the relative change in $y$, $\delta y$ given a small change in $x$, $\delta x$.
However, in today's lesson on partial derivative, my professor constantly used this notation. 
Given a function $z = f(x,y)$, the first derivative with respected to $x$ is written as
$$ \frac{\partial z}{\partial x} $$
So, for example
$$
z = 5x+3y\\
\frac{\partial z}{\partial x} = 5
$$
Why can't I just write it as
$$
z = 5x+3y\\
\frac{d z}{d x} = 5
$$
Is it some convention or am I not understanding something in the notation?

REPLY [16 votes]: A nicer notion is that of the differential:
$$ \text{If} \qquad z = 5x + 3y \qquad  \text{then} \qquad  dz = 5\, dx + 3\,dy $$
Then if you decide to hold $y$ constant, that makes $dy = 0$, and you have $dz = 5 \, dx$.

Another notation that works well with function notation is that if we define
$$ f(x,y) = 5x + 3y$$
then $f_i$ means derivative of $f$ with respect to the $i$-th entry; that is
$$ f_1(x,y) = 5 \qquad \qquad f_2(x,y) = 3 $$
This doesn't work well with a common abuse of notation, though; sometimes people write $f(r,\theta)$ when they really mean "evaluate $f$ at the $(x,y)$ pair whose polar coordinates are $(r, \theta)$" rather than the 'correct' meaning of that expression "evaluate $f$ at $(r, \theta)$". So if you're in the habit of doing that, don't try to indicate derivatives by their position.

I confess I really dislike partial derivative notation; when one writes $\partial/\partial x$, one "secretly" means that they intend to hold $y$ constant, then when one passes it through the differential, one gets
$$ \frac{\partial z}{\partial x} = 5 \frac{\partial x}{\partial x} + 3 \frac{\partial y}{\partial x} = 5 \cdot 1 + 3 \cdot 0 = 5$$
However, the suggestive form of Leibniz notation starts becoming very misleading at this point; for example, let's compute other partial derivatives.

$\partial z / \partial x = 5$, holding $y$ constant as the notation suggests
$\partial x / \partial y = -3/5$, holding $z$ constant as the notation suggests
$\partial y / \partial z = 1/3$, holding $x$ constant as the notation suggests

Then putting it together,
$$ \frac{\partial z}{\partial x} \frac{\partial x}{\partial y} \frac{\partial y}{\partial z} = 5 \cdot \left(-\frac{3}{5}\right) \cdot \frac{1}{3} = -1 $$
This is a big surprise if you expect partial derivatives to behave similarly to fractions as their notation suggests!!!