Partial derivative: taking out the 'f'

[unscientific]

Homework Statement

In the first paragraph, I know its missing a function which they did not put, g. Without puting ∂g/∂x but simply putting ∂/∂x, is that equation even mathematically correct? I know they are "filling in the g later" but does this corrupt the in-between steps in anyway?

In the second paragraph, from step 1 to step 2:

-how can they simply " take out the f "?

-how can they simply replace 'f' by 'g'? would this imply that f = g? (which is obviously not the case)

The Attempt at a Solution

I have come to several conclusions to make sense of what's happening:

1. The term ' ∂/∂x ' by itself is meaningless, only when you slap it together with a function like in (∂/∂x)(∂g/∂x) would it mean that you are partially differentiating 'g' twice with respect to 'x'.I only started the topic of partial differentiation today, and the methods presented here seems weird to me...

 

[cepheid]

∂/∂x is not meaningless. It is the symbol for the partial differentiation *operator*. An operator is a thing that *acts* on a function (i.e. performs some mathematical operation on that function). In this case, the operator performs the operation of taking the partial derivative with respect to x. Therefore, ∂f/∂x and (∂/∂x)f both mean the same thing: that the partial differentiation operator is acting on the function f to produce the partial derivative of f with respect to x.

I don't know why f is suddenly changed to g. Ask your prof.

 

[cepheid]

To elaborate: the first line of your text, which expresses what the ∂/∂x operator becomes in spherical coordinates, will be true *regardless* of what function you plug in. Differentiating wrt x always corresponds to differentiating wrt rho and multiplying by cos(phi) and then subtracting blah blah blah etc etc. This is the power of generalizing to differentiation operators in the first place.

 

[unscientific]

To elaborate: the first line of your text, which expresses what the ∂/∂x operator becomes in spherical coordinates, will be true *regardless* of what function you plug in. Differentiating wrt x always corresponds to differentiating wrt rho and multiplying by cos(phi) and then subtracting blah blah blah etc etc. This is the power of generalizing to differentiation operators in the first place.

Thanks! I think i better understand the meaning of operators now, the phrase " it is used to act on another function" really explains it. The original question was to change change d²f/dy² and d²f/dx² into terms of p, ∅. I suppose by subsituting x(p,∅) and y(p,∅) into f it can be rebranded into a new function g..

Haha I'm not in college at the moment, i finished high school 3 years ago and will enter college in october for a physics degree :)

 

[Mute]

The original question was to change change d²f/dy² and d²f/dx² into terms of p, ∅. I suppose by subsituting x(p,∅) and y(p,∅) into f it can be rebranded into a new function g.

I think that reasoning is most likely correct. The author is being notationally exact and writing [itex]f(x,y) = f(x(p,\phi),y(p,\phi)) \equiv g(p,\phi)[/itex], rather than doing what most people do and just changing notation so that [itex]f(x,y) \rightarrow f(p,\phi)[/itex].