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Parametric Derivative Derivation

  1. Nov 20, 2005 #1
    How does one conclude that [tex] \frac{d^{2} y}{dx^{2}} = \frac{dy\'/dt}{dx/dt} [/tex]?

    Thanks
     
  2. jcsd
  3. Nov 20, 2005 #2

    cronxeh

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  4. Nov 22, 2005 #3
    [tex]\frac{d^2 y}{dx^2} = \frac{d \dot{y'}}{d \dot{x}}[/tex]
    [tex]= \frac{dy'}{dx} = \frac{dy}{dx^2}[/tex]

    Which is clearly incorrect.
     
    Last edited: Nov 22, 2005
  5. Nov 22, 2005 #4

    HallsofIvy

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    And, you will notice that what you wrote was incorrect. What is true is that [tex]\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex] not the second derivative.
    That follows from the chain rule.
     
  6. Nov 22, 2005 #5
    I didn't write
    . I wrote [tex]\frac{d^{2}y}{dx^{2}}= \frac{\frac{dy'}{dt}}{\frac{dx}{dt}}[/tex]
     
  7. Nov 24, 2005 #6
    Yes. Oops.
    [tex]\frac{d^2 y}{dx^2} = \frac{d \dot{y'}}{d \dot{x}}[/tex]
    [tex]= \frac{dy'}{dx} = y''[/tex]
     
  8. Nov 25, 2005 #7

    HallsofIvy

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    Then is y' the derivative with respect to x or t? If with respect to t, then your equation is wrong. If with respect to x, then you are back to the first derivative case.
     
  9. Nov 25, 2005 #8
    [tex]y'' = \frac{d^2 y}{dx^2}[/tex]
     
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