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December 20, 2014

December 20, 2014

Posted by **monic** on Friday, February 8, 2013 at 5:40pm.

f(x) = 8 − x^ex/x + ex

f '(x) =

- calculus 1 -
**Steve**, Friday, February 8, 2013 at 6:09pmHard to tell just what you mean there. If you mean

f(x) = (8-xe^x)/(x+e^x)

f' = [(-e^x - xe^x)(x+e^x) - (8-xe^x)(1+e^x)]/(x+e^x)^2

= (-xe^x - x^2 e^x - e^2x - xe^2x - 8 + xe^x - 8e^x + xe^2x)/(x+e^x)^2

= (e^x(-x+x^2-e^x-xe^x+x-8+xe^x)-8)/(x+e^x)^2

= (e^x(x^2-e^x-8)-8)/(x+e^x)^2

doesn't get much simpler

If I got the original wrong, fix it and visit wolframalpha.com

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