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Homework Help: Derivative notation

  1. Mar 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Is it correct to say that if "y" is a function of x, then

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

    ??
     
  2. jcsd
  3. Mar 15, 2010 #2
    Yes (assuming x is your only variable).
     
  4. Mar 15, 2010 #3

    Mark44

    Staff: Mentor

    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.
     
  5. Mar 15, 2010 #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.
     
  6. Mar 15, 2010 #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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