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December 19, 2014

December 19, 2014

Posted by **Mat** on Monday, May 20, 2013 at 10:14pm.

- Calculus -
**Steve**, Tuesday, May 21, 2013 at 11:42amAssuming the usual carelessness with parentheses, I gather you mean

f(x) = ((x-3)/(2x+3))^2

If f = u^2, f' = 2u u'

So, f' = 2(x-3)/(2x+3) * ((x-3)/(2x+3))'

= 2(x-3)/(2x+3) * [(2x+3)-2(x-3)/(2x+3)^2]

= 2(x-3)/(2x+3)*(9/(2x+3)^2)

= 18(x-3)/(2x+3)^3

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