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Implicit differentiation

  1. Jan 15, 2005 #1
    Say we have two functions of x, v and y, such that v=x+y. How can I find v'?
     
  2. jcsd
  3. Jan 15, 2005 #2

    dextercioby

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    Why doncha differentiate its definition wrt to "x"??

    Daniel.

    PS.Tell me what u get.
     
  4. Jan 15, 2005 #3

    HallsofIvy

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    Since v and y are functions of x, I presume by " v' " you mean "the derivative of v with respect to x", i.e. dv/dx.

    Use the chain rule: v= x+ y so dv/dx= dx/dx+ dy/dx= 1+ dy/dx. What dy/dx is depends, of course, on what function y is of x.
     
  5. Jan 16, 2005 #4
    Oh, so it's only dy/dx? Cause I remember my book doing something like (dy/dx)(dv/dx) or something. Thanks HallsofIvy. :smile:

    Another question. How do I find d(e^u dy/dx)/du, where u and y are functions of x?
    My book says:

    [tex]e^u \frac{dy}{dx} + e^u \frac{d^2y}{dx^2} \cdot \frac{dx}{du}[/tex]

    I understand this is the product rule, but where'd the dx/du come from?
     
    Last edited by a moderator: Jan 16, 2005
  6. Jan 16, 2005 #5

    Galileo

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    It comes from the chain rule again:

    [tex]\frac{d}{du} \frac{dy}{dx}=\frac{d^2y}{dx^2}\frac{dx}{du}[/tex]
     
  7. Jan 16, 2005 #6
    Dank je. :smile:
     
  8. Jan 17, 2005 #7

    Galileo

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    Graag gedaan. :biggrin:
     
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