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    question on derivative I wan chek

    question on derivative I wan chek-195731.jpg
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    MHF Contributor chisigma's Avatar
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    The computation is easier if You take into account the identity...

    \displaystyle \frac{\sqrt{s}-1}{\sqrt{s}+1} = 1 - \frac{2}{\sqrt{s} +1}

    Kind regards

    \chi \sigma
    Last edited by chisigma; December 28th 2010 at 06:08 AM. Reason: corrected a mistake...
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    Is my answer no correct or not ?
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    No, it isn't.

    You have the formula \left(\frac{u}{v}\right)'= \frac{u'v- uv'}{v^2} correct but then you used \frac{uv'- u'v}{v^2} which is wrong.
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  5. #5
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    Quote Originally Posted by chisigma View Post
    The computation is easier if You take into account the identity...

    \displaystyle \frac{\sqrt{s}-1}{\sqrt{s}+1} = 1 - \frac{3}{\sqrt{s} +1}

    Kind regards

    \chi \sigma
    Actually it's \displaystyle \frac{\sqrt{s} - 1}{\sqrt{s} + 1} = 1 - \frac{2}{\sqrt{s} + 1}.
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