Dy/dx - Fraction and/or Operator

You are basically asking the same question as the previous person. The answer is the same- the derivative is a limit of the difference quotient, and the differential is defined in terms of the derivative. So when you "go back" to the difference quotient, you can multiply through by the differentials, and then take the limit to get the derivative back. As far as a "good book" goes... have you tried looking at the thread? I mean, obviously you haven't, because the answer is right there. But if you don't like that one, try Spivak's "Calculus on Manifolds". It's short.In summary, the derivative operator dy/dx
  • #1
p75213
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0
I am a bit confused over the use of the derivative operator dy/dx. I realize 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[itex]\frac{di}{dt}[/itex]
di=[itex]\frac{1}{L}[/itex]v dt
[itex]\int[/itex]di=[itex]\frac{1}{L}[/itex][itex]\int[/itex]v(t) dt

Everything works out nicely but it is a bit confusing when operators can be treated as fractions.
 
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  • #2
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...

[tex]v(t) = L \frac{di}{dt}[/tex]

[tex]\frac{1}{L} v(t) dt = \frac{di}{dt} dt[/tex]

[tex]\int \frac{1}{L} v(t) dt = \int \frac{di}{dt} dt[/tex]

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

[tex]\frac{1}{L} \int v(t) dt = \int di[/tex]
 
  • #3
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
p75213 said:
I am a bit confused over the use of the derivative operator dy/dx. I realize 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[itex]\frac{di}{dt}[/itex]
di=[itex]\frac{1}{L}[/itex]v dt
[itex]\int[/itex]di=[itex]\frac{1}{L}[/itex][itex]\int[/itex]v(t) dt

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

Bacle said:
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.

The differential operator is usually written as d, as in d(t2) = 2t dt.
 
  • #5
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
I think I have figured it out?
y=2x
[itex]\frac{dy}{dx}[/itex]=[itex]\frac{d}{dx}[/itex]2x=2
 
  • #7
p75213 said:
I think I have figured it out?
y=2x
[itex]\frac{dy}{dx}[/itex]=[itex]\frac{d}{dx}[/itex]2x=2

Looks right. The resulting derivative, at least.
 
  • #8
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.
 
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  • #9
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
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.
[itex]\downarrow[/itex]
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
c.teixeira said:
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.
[itex]\downarrow[/itex]
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!

I like HallsofIvy's explanation. Its a couple of posts above
 
  • #12
c.teixeira said:
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.
[itex]\downarrow[/itex]
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!

Have you tried:

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

FAQ: Dy/dx - Fraction and/or Operator

What is Dy/dx?

Dy/dx is a mathematical notation used to represent the derivative of a function. It is commonly used in calculus to indicate the rate of change of a dependent variable (y) with respect to an independent variable (x).

How do you calculate Dy/dx?

To calculate Dy/dx, you need to find the derivative of the function with respect to x. This can be done by using the rules of differentiation, such as the power rule, product rule, quotient rule, and chain rule. Once you have the derivative, you can substitute x into the resulting equation to find the value of Dy/dx.

What is the difference between a fraction and an operator in Dy/dx?

In Dy/dx, a fraction represents the ratio of the change in the dependent variable (y) to the change in the independent variable (x). An operator, on the other hand, is a symbol that indicates a specific operation, such as differentiation. In the notation Dy/dx, the d/dx is the operator, and the fraction represents the derivative.

Why is Dy/dx important in math and science?

Dy/dx is important because it allows us to understand the rate of change of a function. This is crucial in fields such as physics, engineering, and economics, where we need to analyze how variables change over time. It also helps us solve optimization problems and understand the behavior of complex systems.

How is Dy/dx used in real life applications?

Dy/dx has many real-life applications, including predicting stock market trends, designing efficient transportation routes, and optimizing chemical reactions. It is also used in fields like medicine to understand how drugs affect the body over time. Additionally, Dy/dx is essential in engineering for designing and analyzing structures and systems.

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