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Implicit Differentation Help!

  1. May 19, 2005 #1
    Implicit Differentiation...Help!


    Find [tex]\frac{dy}{dx}[/tex] by Implicit Differentiation:
    [tex]\tan(x - y) = \frac{y}{1 + x^2}[/tex]

    [tex]\frac{d}{dx} (\tan(x - y)) = \frac{d}{dx} \left( \frac{y}{1 + x^2} \right)[/tex]
    [tex]\sec^2 (x - y) \cdot \left( 1 - \frac{dy}{dx} \right) = \frac{(1 + x^2) \frac{dy}{dx} - y \frac{d}{dx} (1 + x^2)}{(1 + x^2)^2} = \frac{(1 + x^2) \frac{dy}{dx} - 2xy}{(1 + x^2)^2}[/tex]

    [tex]\sec^2 (x - y) \cdot \left( 1 - \frac{dy}{dx} \right) = \frac{(1 + x^2) \frac{dy}{dx} - 2xy}{(1 + x^2)^2}[/tex]

    What am I doing wrong?

    Any suggestions?...
     
    Last edited: May 19, 2005
  2. jcsd
  3. May 19, 2005 #2
    nothing.... what's the problem?
     
  4. May 19, 2005 #3
    Identity Crisis...



    How do I isolate [tex]\frac{dy}{dx}[/tex] from this identity?

    [tex]\sec^2 (x - y) \cdot \left( 1 - \frac{dy}{dx} \right) = \frac{(1 + x^2) \frac{dy}{dx} - 2xy}{(1 + x^2)^2}[/tex]

    Any suggestions?...
     
  5. May 19, 2005 #4

    Pyrrhus

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    Get together the terms with dy/dx and then factor it. Are you having problem doing that? Distribute (1+x^2)^2 for each term on the right side.
     
  6. May 19, 2005 #5

    dextercioby

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    Implicite differentiation doesn't exclude (the use of) the theorem of implicit functions.So go ahead and use it.

    Daniel.
     
  7. May 20, 2005 #6
    Orion Cheat Theorem...



    Instead of using the Implicit Function Theorem or explicit expansion or distribution, I am wondering if the Orion Cheat Theorem works...?

    Orion Cheat Theorem:

    Identity:
    [tex]\sec^2 (x - y) \cdot \left( 1 - \frac{dy}{dx} \right) = \frac{(1 + x^2) \frac{dy}{dx} - 2xy}{(1 + x^2)^2}[/tex]

    Is replaced by:
    [tex]a(1 - x) = \frac{bx - c}{d}[/tex]

    Then isolated:
    [tex]ad(1 - x) = bx - c[/tex]
    [tex]ad - adx = bx - c[/tex]
    [tex]-adx - bx = -ad - c[/tex]
    [tex]x(-ad - b) = -ad - c[/tex]
    [tex]x = \frac{-ad - c}{-ad - b} = \frac{ad + c}{ad + b}[/tex]
    [tex]x = \frac{ad + c}{ad + b}[/tex]

    Then reconverted:
    [tex]\boxed{\frac{dy}{dx} = \frac{\sec^2 (x - y)(1 + x^2)^2 + 2xy}{\sec^2 (x - y)(1 + x^2)^2 + (1 + x^2)}}[/tex]

    What do you think of the Orion Cheat Theorem? :biggrin:

    What do you think of my solution? :rolleyes:
     
  8. May 20, 2005 #7
    It was good, but, unfortunately its incorrect, because sec^2 isnt a variable, but a function. I like your way of thinking. The easiest way to do this problem is to multiply out the denominator of the RHS on both sides, then add the 2xy to both sides, and divide by whats in the big parentheses.
     
  9. May 20, 2005 #8

    dextercioby

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    It doesn't matter,the notation is formal.That's all there is to it.

    Daniel.
     
  10. May 21, 2005 #9
    What are you referring to? If your saying what he did is fine, I'd have to disagree. What he did is fine for the most part, except he distributed the secant function as if it were a variable, which is clearly incorrect.

    edit: I'm sorry, I misread your translations. It looks like the solution is correct.
     
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