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Math Help - Derivative

  1. #1
    Member
    Joined
    Mar 2010
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    150

    Derivative

    Hello All,

    I've calculated the following answer:

    \frac{-1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}

    And here is the answer in the back of the book:

    \frac{-1}{(1+x)\sqrt{1-x^2}}

    Can someone please help show me how to get from A to B? Thanks
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  2. #2
    MHF Contributor
    Joined
    Nov 2008
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    France
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    Hi

    \frac{-1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}} = \frac{-1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2(1+x)\sqrt{1+x}}

    \frac{-1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}  =  \frac{-(1+x)}{2(1+x)\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}\sqrt{1-x}}{2(1+x)\sqrt{1+x}\sqrt{1-x}}

    \frac{-1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}   =  \frac{-(1+x)}{2(1+x)\sqrt{1+x}\sqrt{1-x}}-\frac{1-x}{2(1+x)\sqrt{1+x}\sqrt{1-x}}

    \frac{-1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}    =  \frac{-(1+x)-(1-x)}{2(1+x)\sqrt{1+x}\sqrt{1-x}}

    \frac{-1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}     =  \frac{-2}{2(1+x)\sqrt{(1+x)(1-x)}}

    \frac{-1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}      =  \frac{-1}{(1+x)\sqrt{1-x^2}}
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  3. #3
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    Mar 2010
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    Thank You very much, I know that must have been a lot of typing!
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