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July 30, 2014

July 30, 2014

Posted by **Brett** on Sunday, November 6, 2011 at 10:44pm.

y= 9e^(x)/ 2e^(x)+ 1

Please Show Work

- Calculus -
**Steve**, Monday, November 7, 2011 at 1:02pmI assume that's 9e^x/(2e^x + 1)

Otherwise, y = 9/2 + 1 = 11/2, which is not very interesting.

y' = [9e^x (2e^x + 1) - 9e^x * 2e^x]/(2e^x + 1)^2

y' = [18e^(2x) + 9e^x - 18e^(2x)]/(2e^x + 1)^2

y' = 9e^x/(2e^x + 1)^2

Hmm. How about that?

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