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January 31, 2015

January 31, 2015

Posted by **Tim** on Sunday, March 16, 2014 at 4:12pm.

y=(e^-xcos^2(x))/(x^2+x+1)

- Calculus Help -
**Steve**, Sunday, March 16, 2014 at 4:35pmlog y = log (e^-x - xcos^2(x)) - log(x^2+x+1)

1/y y' = (-e^-x - (cos^2 x - 2xcosx*sinx))/(e^x - xcos^2 x) - (2x+1)/(x^2+x+1)

y' = (-e^x - cos^2 x + xsin2x) - (2x+1)/(x^2+x+1)^2

If you want to put it all over the common denominator, it gets fairly messy on top.

- Calculus Help -
**Steve**, Sunday, March 16, 2014 at 4:36pmactually, I see I did forget to include a factor of 1/(x^2+x+1) in the first term. Dang!

Luckily, it's not my job to fix it!

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