# Leibniz notation

1. Nov 11, 2013

### BMW

I never really understood leibniz notation. I know that dy/dx means differential of y with respect to x, but what do the 'd's mean? How come the second-order differential is d2y/dx2? What does that mean? And what does d/dx mean?

2. Nov 11, 2013

### Staff: Mentor

The d's stand for "differential". dy/dx is the derivative, not differential, of y with respect to x. The symbol d2y/dx2 represents the derivative (with respect to x) of the derivative with respect to x, or in other words, the 2nd derivative of y with respect to x. It could be written as
$$\frac{d}{dx}(\frac{dy}{dx})$$

As a notational shortcut, the above is often written as d2y/dx2.

d/dx is the differentiation operator, indicating that we're interested in taking the derivative (with respect to x) of whatever function is to the right of this operator.

3. Nov 11, 2013

### BMW

So the second derivative should really be d2y/(dx)2?

4. Nov 11, 2013

### BMW

If you have a differential equation with variables separated, such as dy/dx = 4x2/3y3, and you rearrange it to 3y3 dy = 4x2 dx, what does the dy/dx mean in this case, and can you even rearrange it like that or must you do this: ∫3y3 dy = ∫4x2 dx ?

5. Nov 11, 2013

### Staff: Mentor

Yes.

6. Nov 11, 2013

### Staff: Mentor

It means just what you would think it means - the derivative of y with respect to x.
The assumption here is that there is some differentiable function of x that is the solution to the differential equation. y represents that function.
I don't understand your question. The whole idea of separation of variables is to get all the expressions with y and dy on one side of an equation and all the expressions with x and dx on the other side - then integrate.

If it bothers you that you're integrating with respect to y on one side, but with respect to x on the other, you could think of the left side as being ∫3y3 dy/dx * dx, which simplifies to what you have above.

Possibly you've been told that dy/dx shouldn't be thought of as a fraction. Nevertheless, it's convenient to do so in many cases, such as in separating differential equations.

7. Nov 11, 2013

### iRaid

It's probably this, when I took calculus in highschool the teacher told us that this is an illegal operation.

8. Nov 12, 2013

### BMW

Sorry I should have explained it better. When you rearrange the equation, you get the x and dx on one side, and the y and dy on the other side. Do you have to make each side an integral? E.g. does it have to be ∫x dx = ∫y dy, or can you rearrange to x dx = y dy? If you can, what does the equation mean? What does multiplying x by dx do?

It seems weird to me that the dx and dy somehow magically fit into the integral (e.g. the dx which was part of a ratio now just tells you to integrate with respect to x). Does the dx on one side actually represent some quantity? Or is it more of a concept?

Last edited: Nov 12, 2013
9. Nov 12, 2013

### HallsofIvy

Staff Emeritus
If you mean you have the derivative, dy/dx= f(x), and you go to "dy= f(x)dx" by multiplying both sides by dx, yes, that, "multiplying by dx", is an "illegal operation" specifically because the derivative is NOT a fraction it is NOT "dy divided by dx". But, using the definition of the "anti-derivative" or "integral", we can go to $\int dy= \int f(x)dx$.

It is because of the fact that, while the derivative is NOT a fraction, it can always be treated like one, that we define the "differentials", dy and dx separately so that, once we have that definition, we can think of "dy/dx" as a fraction. It is a "mixed notation" that can be confusing, but useful.

10. Nov 12, 2013

### lavinia

Rephrasing a little what has already been said.

As an equation among differentials, dy = f(x)dx is correct.
How ever dy and dx are not quantities. They are differentials of the functions y and x.

Differentials of functions are what get integrated, not quantities.

In Physics, the expression dy = f(x) dx is taken to mean that for very small Δy and Δx,
the equation Δy = f(x)Δx is approximately true and this approximation gets arbitrarily accurate for smaller and smaller Δx. In fact, Δy/Δx approaches f(x) arbitrarily closely as well. This approximation is expressed as the ratio of infinitesimals dy/dy =f(x) which I would not be surprised actually meant something to Leibniz but nowsdays is taken merely as notation.

Last edited: Nov 12, 2013
11. Nov 12, 2013

### Staff: Mentor

If y = f(x), then your first equation should be dy = f'(x)dx, and the second would be Δy = f'(x)Δx
Δy/Δx approaches f'(x)
dy/dx = f'(x)