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Math Help - Derive arctangent

  1. #1
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    Derive arctangent

    I want to derive this equation:

     <br />
y*arctan(x-2) + x*arctan(\frac{y-1}{20}) = 0<br />

    should i first use the product formulae on the terms on each side of the addition sign? Later should I use the sum formulae on that?
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    You have to use implicit differentiation, the product rule and

    (\arctan{u})'=\frac1{1+u^2}\cdot{u'}
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    You have to use implicit differentiation, the product rule and

    (\arctan{u})'=\frac1{1+u^2}\cdot{u'}
    yea and what happens to the y and the x before the arctan term?
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  4. #4
    Senior Member tukeywilliams's Avatar
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    Quote Originally Posted by wizzler View Post
    I want to derive this equation:

     <br />
y*arctan(x-2) + x*arctan(\frac{y-1}{20}) = 0<br />

    should i first use the product formulae on the terms on each side of the addition sign? Later should I use the sum formulae on that?



    You use the product rule.

    So   \frac{y}{1+ (x-2)^2} + \arctan(x-2)y' + \frac{xy'}{20 \left(1+\left(\frac{y-1}{20} \right)^2 \right)} + \arctan \left(\frac{y-1}{20} \right)
    Last edited by tukeywilliams; August 1st 2007 at 11:24 PM.
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  5. #5
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    earboth's Avatar
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    Quote Originally Posted by tukeywilliams View Post
    You use the product rule.

    So   \frac{y}{1+ (x-2)^2} + \arctan(x-2)y' + \frac{xy'}{20 \left(1+\frac{y-1}{20} \right)^2} + \arctan \left(\frac{y-1}{20} \right)
    Hello,

    I don't want to pick at you ... but I can't see you you around here so I'm going to remove the small typo in your result:

      \frac{y}{1+ (x-2)^2} + \arctan(x-2)y' + \frac{xy'}{20 \left(1+\left(\frac{y-1}{20} \right)^2\right)} + \arctan \left(\frac{y-1}{20} \right)
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