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Thread: derivative of arctan2x^1/2

  1. November 21st 2011, 01:57 PM #1
    Barthayn
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    derivative of arctan2x^1/2

    [QUOTE=Quacky;694202]Let's see.

    y = \sqrt{arctan2x}
    y^2 = arctan2x
    tany^2 = 2x
    \frac {dy}{dx} * sec^2y^2 tany^2 = 2

    How do I differentiate from here? I get \frac {dy}{dx} = \frac {1}{4x^3+x} as an answer
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  2. November 21st 2011, 02:04 PM #2
    e^(i*pi)
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    Re: derivative of arctan2x^1/2

    [QUOTE=Barthayn;696842]
    Quote Originally Posted by Quacky View Post
    Let's see.

    y = \sqrt{arctan2x}
    y^2 = arctan2x
    tany^2 = 2x
    \frac {dy}{dx} * sec^2y^2 tany^2 = 2

    How do I differentiate from here? I am lost because of the y^2
    Use implicit differentiation and the chain rule.

    The derivative of \tan(y^2) is 2y \sec^2(y^2) \cdot \dfrac{dy}{dx}
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  3. November 21st 2011, 02:07 PM #3
    skeeter
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    Re: derivative of arctan2x^1/2

    let u = 2x

    y = \left(\arctan{u}\right)^{1/2}

    y' = \frac{1}{2}\left(\arctan{u}\right)^{-1/2} \cdot \frac{u'}{1+u^2}
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  4. November 21st 2011, 02:16 PM #4
    Barthayn
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    Re: derivative of arctan2x^1/2

    thanks, I got it now
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