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Thread: derivative of arctan2x^1/2

  1. #1
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    derivative of arctan2x^1/2

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

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

    How do I differentiate from here? I get $\displaystyle \frac {dy}{dx} = \frac {1}{4x^3+x}$ as an answer
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  2. #2
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    Re: derivative of arctan2x^1/2

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

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

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

    The derivative of $\displaystyle \tan(y^2)$ is $\displaystyle 2y \sec^2(y^2) \cdot \dfrac{dy}{dx}$
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  3. #3
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    Re: derivative of arctan2x^1/2

    let $\displaystyle u = 2x$

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

    $\displaystyle y' = \frac{1}{2}\left(\arctan{u}\right)^{-1/2} \cdot \frac{u'}{1+u^2}$
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  4. #4
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    Re: derivative of arctan2x^1/2

    thanks, I got it now
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