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Dividing dy/dx by (dy/dx)^2

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data
    Am I right to treat dy/dx as a fraction in this scenerio?

    2. Relevant equations



    3. The attempt at a solution


    [tex]= \frac{\frac{-dy}{dx}} {(\frac{dy}{dx})^2}[/tex]
    [tex]= \frac{\frac{-dy}{dx}} {\frac{dy}{dx} (\frac{dy}{dx})}[/tex]
    [tex]= \frac{1}{\frac{dy}{dx}}[/tex]

    Also, would we be able to multiple out dy/dx if we had for example:

    [tex] 2 = \frac{1}{\frac{dy}{dx}}[/tex]
    [tex] 2(\frac{dy}{dx}) = 1 [/tex]
     
  2. jcsd
  3. Nov 12, 2011 #2

    Mark44

    Staff: Mentor

    Sure.
    What happened to the minus sign?

    Note that you can simplify 1/(dy/dx).
    Sure. And if the goal is to solve for dy/dx, divide both sides of the last equation by 2.
     
  4. Nov 13, 2011 #3
    Simple mis-type.

    Like this?
    [tex]=\frac{dx}{dy}[/tex]

    Which would read change of x with respect to y.. Hmm, I'm not used to it being like that.

    [tex]=\frac{dx}{dy} = lim_{Δx→0} \frac{Δx}{Δy}[/tex]
    [tex]= lim_{Δx→0} \frac{Δx}{f(x+Δx) - f(x)}[/tex]
    [tex]= f'(y)[/tex]

    Is what I stated correct?


    Oh okay, thank you! I just wanted to make sure I wasn't violating any rules here pertaining to derivatives.
     
    Last edited: Nov 13, 2011
  5. Nov 13, 2011 #4

    Mark44

    Staff: Mentor

    No, the limit would be as Δy→0. Here the assumption would be that x is some function of y.
     
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