# Dy/dx - Fraction and/or Operator

1. Sep 27, 2011

### p75213

I am a bit confused over the use of the derivative operator dy/dx. I realise dy is a very small change in y and dx is a very small change in x. When combined into dy/dx it is an operator which means take the derivative of y with respect to x.
However I notice many authors still treat it as a fraction- a small change in y over a small change in x. eg
v=L$\frac{di}{dt}$
di=$\frac{1}{L}$v dt
$\int$di=$\frac{1}{L}$$\int$v(t) dt

Everything works out nicely but it is a bit confusing when operators can be treated as fractions.

2. Sep 27, 2011

### Char. Limit

Strictly speaking, dy/dx cannot be taken as a fraction. However, as an abuse of notation in SOME cases, they will use this as a fraction regardless. Usually, this can be done rigorously another way. For example, if you'll allow multiplication by differentials...

$$v(t) = L \frac{di}{dt}$$

$$\frac{1}{L} v(t) dt = \frac{di}{dt} dt$$

$$\int \frac{1}{L} v(t) dt = \int \frac{di}{dt} dt$$

Then using the u-substitution on the right-hand side, di/dt dt simplifies to di, and we get the problem as originally stated.

$$\frac{1}{L} \int v(t) dt = \int di$$

3. Sep 27, 2011

### Bacle

I have found a bit more clarity thinking of dy/dx as a _differential_ operator, which assigns to a differentiable f its differential f'(t)dt , which is the local-linear approximation to the change of values of f, but there may be some uses (and maybe abuses) of notation that I am not familiar with.

4. Sep 28, 2011

### Staff: Mentor

The derivative operator is d/dx, not dy/dx. The first symbol operates on a differentiable function of x. The second symbol represents the derivative (with respect to an independent variable x) of a differentiable function y.

The differential operator is usually written as d, as in d(t2) = 2t dt.

5. Sep 28, 2011

### Bacle

You're right, Mark44 , d/dt is the usual format for the operator assigning the
differential . Moreover, the differential of a differentiable function is a differential form.

6. Sep 30, 2011

### p75213

I think I have figured it out?
y=2x
$\frac{dy}{dx}$=$\frac{d}{dx}$2x=2

7. Oct 3, 2011

### MrNerd

Looks right. The resulting derivative, at least.

8. Oct 4, 2011

### HallsofIvy

As Mark44 said, d/dx is the "operator", not dy/dx.

dy/dx is NOT a fraction- but it can be treated like one. Specifically, dy/dx is the limit of the "difference quotient" (f(x+h)- f(x))/h. So you can "go back before the limit", use the appropriate fraction property, and then take the limit.

To make that "treat the derivative as a fraction" rigorous, we define the "differentials" dx and dy separately- though most elementary texts just "hand wave" those definitions.

Last edited by a moderator: Mar 1, 2012
9. Oct 4, 2011

### Bacle

Well, if y(x) is differentiable, then dy is f'(x)dx, and dx is just dx.

dy is the change of y(x) along the tangent line (seen as a limiting position of the secant).

But I agree that a lot of texts on PDE's just happily cross-multiply in cases of separation of

variables, without justification.

10. Feb 29, 2012

### c.teixeira

I am revisiting ODE's, and this doubt is killing me as well.

I am re-learning by ( ODE's ( Tenenbaum and Pollard) from Dover.

It is pretty clear that dy/dx represents f(x), as it is also very easy to understand "geometrically" that dy = f(x) dx.
However, altough I am confortable solving ODE's, I just don't understant how the hell is it mathematically possible to multiply an equation by dx in order to integrate separately.
For example:

Q(x,y) dy/dx + P(x,y) = 0.
$\downarrow$
P(x,y) dx + Q(x,y) dy = 0.

They treat dy/dx as a fraction. I would like to know why, how is that possible.
IN fact I would love if you guys could recommend me a good book that explains this very clearly.
Thanks!

11. Feb 29, 2012

### p75213

I like HallsofIvy's explanation. Its a couple of posts above

12. Feb 29, 2012

### Bacle2

Have you tried:

http://en.wikipedia.org/wiki/Separation_of_variables ?

13. Apr 12, 2012