Relationship between differential and the operator

[wumple]

Is there some sort of relationship between the differential

dx

and the differential operator which means to take the derivative d/dx

if x is a dependent variable? My prof said that dx * d/dx = 1 but that doesn't seem to work out in the case I'm looking at, so I must be missing something.

 

[Char. Limit]

dx \frac{d}{dx} is just an operator multiplied by a differential. If you apply it to a function f(x), it'll give the differential df, however. So you could call it the function differential operator, I suppose.

 

[wumple]

So then can you take dx = 1/(d/dx)? As in the differential is one over the operator? Or does that not work? I was having trouble making that work out in one of my problems.

 

[Char. Limit]

No, that wouldn't work. I really don't believe it's possible to say that dx d/dx = 1, anyway. If you're really trying to assign it a value, it would be just d, or a differential operator. That is, you're basically saying "the derivative operator with respect to a variable, multiplied by the differential of that variable, is the differential operator".

 

[chiro]

You might want to revert back to the limit definition for the derivative for understanding cases like this.

 

[DrewD]

On a multivariable function, \frac{\partial}{\partial x} and \partial x can be thought of as unit vectors. I am still not totally comfortable with the idea, but perhaps this is what your prof was getting at?

 

[mathwonk]

have you had linear algebra? df is a field of (co)vectors, and dx is a field of standard basis covectors. In this basis, d/dx is the operator that sends f to its field of coefficients in terms of the basis dx. I.e. at every point p, df(p) = df/dx(p) dx.

I.e. at every point df/dx(p) is a number, and at every point df(p) is a covector.

At every point dx(p) is also a covector, and we have at every point, that the number df/dx(p) times the covector dx(p), equals the covector df(p).

so df/dx . dx = df, an equation that holds at every point p.for instance if f(x) = x^2, then df(p) = 2p. dx(p) = df/dx(p) dx(p).

 

[schliere]

Is there some sort of relationship between the differential

dx

and the differential operator which means to take the derivative d/dx

if x is a dependent variable? My prof said that dx * d/dx = 1 but that doesn't seem to work out in the case I'm looking at, so I must be missing something.

Your professor is wrong. dx * d/dx = d. In words, the change [of something] per change in x times the change in x is the change [in the thing].