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December 18, 2014

December 18, 2014

Posted by **john** on Thursday, March 31, 2011 at 2:29pm.

- calculus -
**Reiny**, Thursday, March 31, 2011 at 2:33pmI don't know what name they give the rule these days, but the way I do it would get you ...

f'(x) = (3/2)(2x-6)(x^2 - 6x + 23)^(1/2) in one step.

and

= 3(x-3)√(x^2 - 6x + 23)

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