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Math Help - Derivative of arctan...

  1. #1
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    Derivative of arctan...

    I'm currently working this problem


    pretty unsuccessfully. I tried to use the derivative of arctan and the general power rule and as a result I got:

    But this is wrong. Any thoughts? Thanks to anybody that helps.

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  2. #2
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    You should have three segments:

    \frac{1}{1+(4x^2-1)} \cdot \frac{1}{2(\sqrt{4^2-1})} \cdot 8x
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  3. #3
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    Using the chain rule, and the derivative of arctan, we'll get

    \frac{\mathrm{d}}{\mathrm{d} x} (\arctan u)=\frac{{u}'}{1+u^2}

    With u=\sqrt{4x^2-1}

    \frac{\frac{4x}{\sqrt{4x^2-1}}}{1+4x^2-1}=\frac{4x}{4x^2\cdot \sqrt{4x^2-1}}=\frac{1}{x\cdot \sqrt{4x^2-1}}

    So you were right =)
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  4. #4
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    Quote Originally Posted by fleabass123 View Post
    I'm currently working this problem


    pretty unsuccessfully. I tried to use the derivative of arctan and the general power rule and as a result I got:

    But this is wrong. Any thoughts? Thanks to anybody that helps.

    actually, what you arrived at is correct.

    u = \sqrt{4x^2-1}

    u' = \frac{4x}{\sqrt{4x^2-1}}

    \frac{d}{dx} \tan^{-1}(u) = \frac{u'}{1+u^2}

    \frac{4x}{\sqrt{4x^2-1}} \cdot \frac{1}{1 + (4x^2-1)} = \frac{1}{x\sqrt{4x^2-1}}
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  5. #5
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    I see. Thanks for the help, everybody!
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