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Verification of Implicit Differentiation

  1. Jan 18, 2005 #1
    Help with Implicit Differentiation

    Hello all

    If we are given [tex] \cos xy = 2x^2 - 3y [/tex] find [tex] \frac {dy}{dx} [/tex]

    So the derivative of [tex] \cos xy [/tex] is [tex] - sin(xy)(x)(\frac{dy}{dx} + y) [/tex]

    The derivative of the RHS is [tex] 4x - 3 \frac {dy}{dx} [/tex]

    Hence [tex] \-sin(xy)(x)\frac{dy}{dx} + y [/tex] = [tex] 4x - 3 \frac {dy}{dx} [/tex]

    HOw do I find [tex] \frac {dy}{dx} [/tex] ?

    Is the answer [tex] \frac {dy}{dx} = \frac{y\sin xy + 4x}{ -x\sin xy + 3} [/tex]


    ( sin should be (- sin)
    Last edited: Jan 18, 2005
  2. jcsd
  3. Jan 18, 2005 #2
    I think You do not correctly found derivative of left part.
    Hint: (x*y)'=y+x*y'
  4. Jan 18, 2005 #3
    Sorry, I didn't noticed correction. You are right.
  5. Jan 18, 2005 #4
    anybody who can verify if my answer is right?
  6. Jan 18, 2005 #5
    The answer is right. You have tipos
  7. Jan 18, 2005 #6
    I think you have the right idea, but you need to clean up your work,,, for example
    the fifth line should not be
    [tex] -sin(xy)(x)\frac{dy}{dx} + y [/tex]
    you need to distribute the [itex] \-sin(xy) [/itex] to the y as well
    so it should be:
    [tex] -sin(xy)(x\frac{dy}{dx} + y) [/tex] ( I assumed you knew this, and just had difficulty with the syntax )

    Anyway, once you have differentiated implicitly you can treat [itex] \frac{dy}{dx}[/itex] as a your unknown and solve for it algebraicly

    Keep at it :)
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