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Math Help - Find the second derivative of

  1. #1
    Senior Member
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    Find the second derivative of

    y=\frac{\sqrt{1-x^2}}{x}

    This is what I get so far:
    y'=\frac{-x-\sqrt{1-x^2}}{x^2\sqrt{1-x^2}}

    y''= I'm working on it but I'm kinda confused.

    Is y' correct or did I mess up?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Swampscott, MA
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    Quote Originally Posted by yoman360 View Post
    y=\frac{\sqrt{1-x^2}}{x}

    This is what I get so far:
    y'=\frac{-x-\sqrt{1-x^2}}{x^2\sqrt{1-x^2}}

    y''= I'm working on it but I'm kinda confused.

    Is y' correct or did I mess up?
    I hate the quotient rule and will do anything I can to avoid it.

    \frac{\sqrt{1-x^2}}{x}=\frac{\sqrt{1-x^2}}{\sqrt{x^2}}=\sqrt{x^{-2}-1}

    \frac{d}{dx}\left[\sqrt{x^{-2}-1}\right]=\frac{1}{2\sqrt{x^{-2}-1}}\cdot\frac{-2}{x^3}=\frac{-1}{x^3\sqrt{x^{-2}-1}}=\frac{-1}{\sqrt{x^4-x^6}}=-(x^4-x^6)^{-1/2}

    \frac{d}{dx}\left[-(x^4-x^6)^{-1/2}\right]=\frac{1}{2}(x^4-x^6)^{-3/2}\cdot(4x^3-6x^5)=\boxed{\frac{2x^3-3x^5}{(x^4-x^6)^{3/2}}}
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  3. #3
    Super Member redsoxfan325's Avatar
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    Actually, you can simplify it even more.

    \frac{2x^3-3x^5}{(x^4-x^6)^{3/2}}=\frac{2x^3-3x^5}{[x^2(x^2-x^4)]^{3/2}}=\frac{2x^3-3x^5}{x^3(x^2-x^4)^{3/2}}=\boxed{\frac{2-3x^2}{(x^2-x^4)^{3/2}}}
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