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The computation is easier if You take into account the identity... $\displaystyle \displaystyle \frac{\sqrt{s}-1}{\sqrt{s}+1} = 1 - \frac{2}{\sqrt{s} +1}$ Kind regards $\displaystyle \chi$ $\displaystyle \sigma$
Last edited by chisigma; Dec 28th 2010 at 06:08 AM. Reason: corrected a mistake...
Is my answer no correct or not ?
No, it isn't. You have the formula $\displaystyle \left(\frac{u}{v}\right)'= \frac{u'v- uv'}{v^2}$ correct but then you used $\displaystyle \frac{uv'- u'v}{v^2}$ which is wrong.
Originally Posted by chisigma The computation is easier if You take into account the identity... $\displaystyle \displaystyle \frac{\sqrt{s}-1}{\sqrt{s}+1} = 1 - \frac{3}{\sqrt{s} +1}$ Kind regards $\displaystyle \chi$ $\displaystyle \sigma$ Actually it's $\displaystyle \displaystyle \frac{\sqrt{s} - 1}{\sqrt{s} + 1} = 1 - \frac{2}{\sqrt{s} + 1}$.
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