I have a clarifying question about this question:
What is the difference between $d$ and $\partial$?
I understand the idea that $\frac{d}{dx}$ is the derivative where all variables are assumed to be functions of other variables, while with $\frac{\partial}{\partial x}$ one assumes that $x$ is the only variable and every thing else is a constant (as stated in one of the answers).
Example 1: If $z = xa + x$, then I would guess that
$$
\frac{\partial z}{\partial x} = a + 1
$$
and
$$
\frac{d z}{d x} = a + x\frac{da}{dx} + 1.
$$
since now $a$ should be considered a function.
When we in calculus 1 have $y = ax^2 + bx + c$, then technically we should use $\partial$ as we are assuming $a, b$, and $c$ are constants?
Is this correct?
Example 2: Maybe the thing that is confusing me is that when we do implicit differentiation we use $d$. So if
$$
x^2 + y^2 = 1
$$
then taking $\frac{d}{dx}$ gives
$$
2x + 2y\frac{dy}{dx} = 0
$$
again because $y$ is considered a function.
How would taking $\frac{\partial}{\partial x}$ of an equation like $x^2 + y^2 =1$ work? Does that even make sense?
Example 3: Is it ever possible that using $\partial$ and $d$ can give the same? If, for example $y = x^2$, does it make sense to say that
$$
\frac{\partial}{\partial x} y = 2x?
$$
Edit: My overall question, I guess, is how the notations of partial derivatives vs. ordinary derivatives are formally defined. I am looking for a bit more background.
Best Answer
Some key things to remember about partial derivatives are:
So for your Example 1, $z = xa + x$, if what you mean by this to define $z$ as a function of two variables, $$z = f(x, a) = xa + x,$$ then $\frac{\partial z}{\partial x} = a + 1$ and $\frac{dz}{dx} = a + 1 + x\frac{da}{dx},$ as you surmised, though you could also have gotten that last result by considering $a$ as a function of $x$ and applying the Chain Rule.
But when we write something like $y = ax^2 + bx + c,$ and we say explicitly that $a$, $b$, and $c$ are (possibly arbitrary) constants, $y$ is really only a function of one variable: $$y = g(x) = ax^2 + bx + c.$$ Sure, you can say that $\frac{\partial y}{\partial x}$ is what happens when you vary $x$ while holding $a$, $b$, and $c$ constant, but that's about as meaningful as saying you vary $x$ while holding the number $3$ constant.
I suppose technically $\frac{\partial y}{\partial x}$ is defined even if $y$ is a single-variable function of $x$, but it would then just be $\frac{dy}{dx}$ (the ordinary derivative), and I can't remember seeing such a thing ever written as a partial derivative. It would not make it possible to do anything you cannot do with the ordinary derivative, and it might confuse people (who might try to guess what other variables $y$ is a function of).
The previous paragraph implies that the answer to your Example 3 is "yes." It also hints at why I almost wrote "a function of two or more variables" as part of the first requirement for using partial derivatives. Technically I think you only need a function of one or more variables, but you should want a function of at least two variables before you think about taking partial derivatives.
For Example 2, where we have $x^2 + y^2 = 1$, it is not obvious what the function is that we would be taking partial derivatives of. Either $x$ or $y$ could be a function of the other. (The function would be defined only over a limited domain, and would produce only some of the points that satisfy the equation, but it can still be useful to do some analysis under those conditions.) If you write something besides the equation to make it clear that (say) $y$ is a function of $x$, giving a sufficiently clear idea which of the possible functions of $x$ you mean, then I think technically you could write $\frac{\partial y}{\partial x}$, and you might even find that $\frac{\partial y}{\partial x} = 2x$, but again this is a lot of trouble and confusion to get a result you could get simply by using ordinary derivatives.
On the other hand, suppose we say that $$h(x,y) = x^2 + y^2 - 1,$$ and we are interested in the points that satisfy $x^2 + y^2 = 1$, that is, where $h(x,y) = 0$. Now we have a function of multiple variables, so we can do interesting things with partial derivatives, such as compute $\frac{\partial h}{\partial x}$ and $\frac{\partial h}{\partial y}$ and perhaps use these to look for trajectories in the $x,y$ plane along which $h$ is constant. OK, we don't really need partial derivatives to figure out that those trajectories will run along circular arcs, but we could have some other two-variable function where the answer is not so obvious.