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October 1, 2014

October 1, 2014

Posted by **Jessi** on Monday, February 20, 2012 at 3:23am.

- Calculus-derivatives -
**drwls**, Monday, February 20, 2012 at 4:39amh(x) = [x/(1-x)]^2 = x^2 * (1-x)^-2

h'(x) = 2x/(1-x)^2 +(-2)(-1)*x^2*(1-x)^-3

= 2x/(1-x)^2 + 2x^2/(1-x)^3

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