Why is there a separate symbol for partial derivatives?

Question

The concept of a partial derivative is very simple: for a multivariate function $f$, the partial derivative of $f$ with respect to a single variable $x$ is computed by treating the other variables as constants and differentiating $f$ with respect to $x$.

As a student of Calculus I, I do not fully understand the need for a $\partial y / \partial x$. As far as I know, the special “partial” $\partial$ does not change the process of the computation. In addition, I have become confused as the calculus of my physics course increases in difficulty.

Take a three-dimensional position vector $\vec r = \vec x + \vec y + \vec z$ and an electric field vector $\vec E$ that varies with $\vec r$. If $V$ denotes electric potential, then $\Delta V = -\int \vec E\cdot d\vec r$.

On a review sheet that my teacher created, he expanded this, saying

$$\begin{align} E_x &= -\partial V / \partial x \\ E_y &= -\partial V / \partial y \\ E_z &= -\partial V / \partial z \end{align}$$

Given the standard setup of a partial derivative, I see no issue with this. However, our standard formula chart reads that

$$E_x = -\frac{dV}{dx}$$

and this genuinely confuses me since our calculations of field potential are almost always expanded to multiple dimensions.

What is the need for a $\partial f / \partial x$ notation, and why are sources (at least in physics, the only application of partial derivatives I encounter during the course of the school day) inconsistent?

Using $\partial$ automatically implies there are more than one variable in the differentiated function. It is just helpful. — KKZiomek, Feb 18 '17 at 06:17
I think the only reasonable way to answer this question is to give an example where using $d$ vs. $\partial$ makes a difference. Unfortunately, I'm having trouble coming up with a "natural" one. None of the answers so far use examples in this way. — Ben Grossmann, Feb 18 '17 at 06:21
@Omnomnomnom I just had a look at the question you have indicated: the simple example given by Hurkyl is striking. — Jean Marie, Feb 18 '17 at 06:59

score 8 · Answer 1 · answered Feb 18 '17 at 06:19

Alan Turing said:

The Leibniz notation $\mathrm dy/\mathrm dx$ I find extremely difficult to understand in spite of it having been the one I understood best once! It certainly implies that some relation between $x$ and $y$ has been laid down e.g. $$ y = x^2 + 3x $$

Whenever the notation $\partial z/\partial x$ is used, there exists some sort of implicit dependency of $z$ w.r.t. $x$, e.g. $z = f(x,y)$. The $\partial$ symbol is used since in this case only the dependency of $z$ w.r.t. $x$ is considered. If there is another implicit dependency $y = g(x)$, $\partial z/\partial x$ does not take that into account (hence the name partial derivative), while $\mathrm dz/\mathrm dx$ does, and it is called the total derivative of $z$ w.r.t. $x$.

I always thought that partial meant that the derivative was partial to a certain direction. — Steven Alexis Gregory, Feb 18 '17 at 07:05
@Henricus V. "If there is another implicit dependency [...]", it would help me understand the motivation if I had an example from physics of a useful function f(x, y), where y = g(x), and the partial was specifically relevant where the total isn't? Are you aware of any? — nmr, Jul 07 '24 at 12:24

score 0 · Answer 2 · answered Feb 18 '17 at 06:19

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There is no one forcing you to use $\partial$ for partial derivatives, however it is recommended because it helps avoid confusion in expressions where both partial and ordinary derivatives are present.

answered Feb 18 '17 at 06:19

KKZiomek

3,923

score 0 · Answer 3 · edited Apr 13 '17 at 12:21

I would just like to add that I've done a bit more research and that the distinction between $\partial t$ and $dt$ seems vital for formulae like

$$d\left( f\left( x,y,\cdots\right)\right) = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \cdots$$

and

$$\begin{align} \frac{dz}{dx} &= \frac{dz}{dy} \frac{dy}{dx} \qquad \mathrm{\left( chain \ rule\right)} \\ \frac{\partial z}{\partial x} &\neq \frac{\partial z}{\partial y} \frac{\partial y}{\partial x} \end{align}$$

like in Hurkyl's example $$\frac{\partial z}{\partial x} \frac{\partial x}{\partial y} \frac{\partial y}{\partial z} = \left( 5\right) \left( -\frac{3}{5}\right) \left( \frac{1}{3}\right) = -1$$ if $z = 5x + 3y$

Why is there a separate symbol for partial derivatives?

3 Answers3