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Math Help - Arctan Derivitive Question

  1. #1
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    Arctan Derivitive Question

    Can someone please help me in finding the derivitive of the function:

    Technically it's finding (dy/dx), but please help!
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  2. #2
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    Rearrange

    \tan y=\sqrt{5x^2-1}\implies y'\sec^2y=\frac{5x}{\sqrt{5x^2-1}}

    Make y' the subject of the equation.
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  3. #3
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    I need the answer in the form " dy/dx = ", and quite frankly I have no idea how to go about doing this. If someone can post it I'd be grateful.
    Last edited by Quiz0n; October 8th 2007 at 11:53 AM.
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  4. #4
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    Hello, Quiz0n!

    You are expected to know that: . \frac{d}{dx}\!\left(\tan^{\text{-}1}\!u\right) \;=\;\frac{1}{1+u^2}\!\cdot\!\frac{du}{dx} . . . and the Chain Rule.

    So exactly where is your difficulty?


    Differentiate: . y \;=\;\tan^{-1}\left(5x^2-1\right)^{\frac{1}{2}}

    We have: . \frac{dy}{dx}\;=\;\frac{1}{1 + \left[\left(5x^2-1\right)^{\frac{1}{2}}\right]^2} \cdot \frac{1}{2}(5x^2-1)^{\text{-}\frac{1}{2}} \cdot 10x \;=\;\frac{1}{1 + 5x^2 - 1}\cdot\frac{5x}{\sqrt{5x^2-1}}

    Therefore: . \frac{dy}{dx}\;=\;\frac{5x}{5x^2\sqrt{5x^2-1}} \;=\;\frac{1}{x\sqrt{5x^2-1}}

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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, Quiz0n!

    You are expected to know that: . \frac{d}{dx}\!\left(\tan^{\text{-}1}\!u\right) \;=\;\frac{1}{1+u^2}\!\cdot\!\frac{du}{dx} . . . and the Chain Rule.

    So exactly where is your difficulty?



    We have: . \frac{dy}{dx}\;=\;\frac{1}{1 + \left[\left(5x^2-1\right)^{\frac{1}{2}}\right]^2} \cdot \frac{1}{2}(5x^2-1)^{\text{-}\frac{1}{2}} \cdot 10x \;=\;\frac{1}{1 + 5x^2 - 1}\cdot\frac{5x}{\sqrt{5x^2-1}}

    Therefore: . \frac{dy}{dx}\;=\;\frac{5x}{5x^2\sqrt{5x^2-1}} \;=\;\frac{1}{x\sqrt{5x^2-1}}

    the poster wants dx/dy
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  6. #6
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    Thank you guys so much, the post above was meant to say " dy / dx ." Sorry for the mistype, Jhevon. Thanks again!
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