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November 27, 2014

November 27, 2014

Posted by **sh** on Thursday, April 29, 2010 at 4:17am.

y=ln(x/(2x+3))^(1/2)

y=(1/2)ln(x/(2x+3))

y'=0+(1/2)((2x+3-2x)/(2x+3)²)

What's next after the quotient rule?

- Calculus -
**drwls**, Thursday, April 29, 2010 at 5:36amYou don't have to use the quotient rule of differentiation.

y = lnx - (1/2)ln(2x+3)

y' = 1/x -(1/2)[1/(2x+3)]*2

= (1/x) - [1/(2x+3)]

which can be rewitten with a common denominator, if you wish.

- Calculus -
**sh**, Thursday, April 29, 2010 at 10:47amThanks for your help!

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