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Math Help - derivative check 2

  1. #1
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    derivative check 2

     <br />
\frac{f(x)}{g(x)} = \frac{\sqrt{x^2+1}cosx}{x} = \frac{(x^2+1)^\frac{1}{2}cosx}{x}<br />

     <br />
f ' (x) = \frac{1}{2}(x^2+1)^\frac{-1}{2}cosx - \sqrt{x^2+1}sinx<br />

    so

     <br />
y ' = \frac{x \frac{1}{2}(x^2+1)^\frac{-1}{2}cosx - \sqrt{x^2+1}sinx - \sqrt{x^2+1}cosx}{x^2}<br />

    correct? thank you
    Last edited by jvignacio; November 21st 2008 at 09:37 AM.
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  2. #2
    Senior Member Twig's Avatar
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    hi

    Hi

    I did not go through all your calculations, but I noticed something at the
    beginning. Can you see what it is?

    \sqrt{x^2 + 1} = (x^2 + 1)^{1/2}

    You wrote: \sqrt{x^2 + 1} = (x^2 + 1)^{-1/2}

    Can you see which one is correct?
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  3. #3
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    Quote Originally Posted by Twig View Post
    Hi

    I did not go through all your calculations, but I noticed something at the
    beginning. Can you see what it is?

    \sqrt{x^2 + 1} = (x^2 + 1)^{1/2}

    You wrote: \sqrt{x^2 + 1} = (x^2 + 1)^{-1/2}

    Can you see which one is correct?
    yes sorry mate i ment 1/2 not -1/2!! but wierd i didnt add it to the next line xD
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