# What's the difference between dx and ∂x?

1. Jun 20, 2013

### izabo

I understand from the equation:
$$df = {\frac{\partial f}{\partial x}} dx + {\frac{\partial f}{\partial y}} dy$$
that: $dx \neq \partial x$.

I understand why $df \neq \partial f$
($df$ is the change in f when the change in all the variables is infinitesimal, and $\partial f$ is the change in f when the change in one vriable is infinitesimal and the others are constant, right?).

but doesn't both $dx$ and $\partial x$ mean an infinitesimal change in x? if so, why aren't they equal? if not what do they mean?

2. Jun 20, 2013

### Staff: Mentor

x could depend on y, for example. Or both x and y could depend on some common other variables. In the same way as $df \neq \partial f$, $dx \neq \partial x$

3. Jun 20, 2013

### izabo

how could x depend on y? aren't they both independent variables?

4. Jun 20, 2013

### Staff: Mentor

Who said that?

5. Jun 20, 2013

### izabo

does it really matter?

btw i meant that for f(x,y) if that wasn't clear.

6. Jun 21, 2013

### Fredrik

Staff Emeritus
If you want to make sense of this for a function $f:\mathbb R^2\to\mathbb R$ (without using definitions from differential geometry), then you should define $df:\mathbb R^4\to\mathbb R$ by
$$df(x,y,z,w)=D_1f(x,y)z+D_2f(x,y)w$$ for all $x,y,z,w\in\mathbb R$. This ensures that for all $x,y,dx,dy\in\mathbb R$, we have
$$df(x,y,dx,dy)=D_1f(x,y)dx+D_2f(x,y)dy.$$ What you wrote can be considered a sloppy notation for this result.

This is a way to make sense of df, dx and dy. I don't think I have ever seen anyone try to make sense of $\partial x$. It's just a small part of the notation $\partial f/\partial x$, which means $D_1f$. I find the notation $\partial f/\partial x$ misleading, since it hides the fact that $D_1f$ (the partial derivative of f with respect to the first variable slot) is the same function no matter what symbols we usually use to represent the numbers we plug into $f$.

Note that with this definition, there's no need for dx and dy to be "infinitesimal". What does that even mean? I know that there's a definition of that term, but I haven't studied it, and I'm pretty sure that almost no one of the authors of physics books who use that term have either. If you see the term "infinitesimal" in a physics book, you should assume that it has nothing to do with infinitesimals. You should interpret it as a code that let's you know that the next thing that follows is a first-order approximation. For example, to say that for infinitesimal x, we have $e^x=1+x$, is to say (in a weird and confusing way) that there's a function $R:\mathbb R\to\mathbb R$ such that $e^x=1+x+R(x)$ for all $x\in\mathbb R$, and $R(x)/x\to x$ as $x\to 0$.

df(x,y,z,w) is a first-order approximation of the difference f(x+z,y+w)-f(x,y). So if we want to be weird and confusing, we can say that for infinitesimal z and w, we have f(x+z,y+w)-f(x,y) = df(x,y,z,w). If we want to cause some additional confusion, we can use the notations dx and dy for the real numbers z and w, and then drop the (x,y,z,w) from df(x,y,z,w), and the (x,y) from $D_1f(x,y)$ and $D_2f(x,y)$. Then we would be saying that for infinitesimal dx and dy, we have
$$f(x+dx,y+dy)-f(x,y)=df=D_1f dy+ D_2f dx =\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy.$$

Last edited: Jun 21, 2013
7. Jul 24, 2016

### Consolacion Ruiz

Δ = difference

d = Δ but small difference, infinitesimal

δ = d but along a curve

Mathematical symbols are always graphics.

I’m not sure if that will be true, but it would be beautiful.