question on derivative I wan chek

• Dec 27th 2010, 11:25 PM
r-soy
question on derivative I wan chek
• Dec 28th 2010, 12:33 AM
chisigma
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

$\displaystyle \chi$ $\displaystyle \sigma$
• Dec 28th 2010, 01:18 AM
r-soy
Is my answer no correct or not ?
• Dec 28th 2010, 02:34 AM
HallsofIvy
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.
• Dec 28th 2010, 03:04 AM
Prove It
Quote:

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