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February 27, 2015

February 27, 2015

Posted by **Anonymous** on Sunday, October 7, 2012 at 11:21pm.

- Math -
**Steve**, Monday, October 8, 2012 at 9:55amy = √x/(3x+1)

y' = (1/2√x (3x+1) - √x (3)) / (3x+1)^2

= (3x+1) - 2√x√x(3)) / 2√x (3x+1)^2

= (1-3x) / 2√x (3x+1)^2

Looks like you forgot a factor of √x:

y = u/v

y' = (u'v -**u**v')/v^2

- Math -
**Steve**, Monday, October 8, 2012 at 9:57amalso ,my own typing was sloppy.

y' = (1/(2√x) (3x+1) - √x (3)) / (3x+1)^2

Looks like you forgot to put the numerator over a common denominator of 2√x

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