Derivative notation

  • #1

Homework Statement


Is it correct to say that if "y" is a function of x, then

[tex]y'=\frac{dy}{dx}[/tex]

??
 

Answers and Replies

  • #2
225
0
Yes (assuming x is your only variable).
 
  • #3
34,876
6,615
Those are two different ways of writing one thing.

The y' notation is sometimes called Newton notation (although he used a dot instead of a prime), and the other is called Leibniz notation.
 
  • #4
I think I may have answered my own question with this next one, but I would like to get a confirmation.

I asked the question above because in my math methods in physics class, the tutorials we use and the professor often algebraically manipulate the dy's and dx's in equations involving y'.

I am reviewing scale invariance and its applications to FODE's.

After proving the following equation is scale invariant, I am to solve for the general solution.

[tex]y'+\frac{y^{2}}{x^{2}}=2[/tex]

After I have shown it is scale invariant, I use the substitution y=vx to obtain:

[tex]v+x\frac{dv}{dx}+v^2=2[/tex]

In order to get to the point above, I note that:

[tex]dy=vdx+xdv[/tex]

Here is where my question comes in. I originally looked at this problem by stating:

[tex]y'=\frac{dy}{dx}=\frac{d(vx)}{dx}=\frac{dvdx}{dx}=dv[/tex]

This is the wrong simplification

As far as I know, v is a number not a function of x. I believe I understand that this method is used to make an inhomogeneous equation separable. How do I avoid making the wrong simplification?

I think the answer is that I need to recognize that I must have a "dv" and a "dx" in order to solve the equation in a separable manner, otherwise I would have had:

[tex]dv+v^2=2[/tex]

Does that make any sense at all? Thanks for replying.
 
  • #5
Haha...nevermind. It is the same thing, just not carried out all the way through.

dy/dx=dv........dy=dv(dx)= vdx+xdv.
 

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