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April 18, 2015

April 18, 2015

Posted by **Maho** on Tuesday, March 18, 2014 at 10:47pm.

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

- Calculus Help -
**Reiny**, Tuesday, March 18, 2014 at 11:12pmy = (e^-x) (cosx)^2 (x^2 + x + 1)

take ln of both sides

ln y = ln e^-x + ln (cosx)^2 + ln(x^2 + x + 1)

= -x + 2 ln(cosx) + ln(x^2 + x + 1)

now differentiate

y' / y = -1 + 2(-sinx/cosx) + (2x+1)/(x^2 + x + 1)

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

y' = y(-1 - 2tanx + (2x+1)/(x^2 + x + 1))

= [ (e^-x) (cosx)^2 (x^2 + x + 1) ] * [ -1 - 2tanx + (2x+1)/(x^2 + x + 1) ]

sure hope they don't expect us to simplify this

- Calculus Help -
**Steve**, Tuesday, March 18, 2014 at 11:15pmlog y = log (e^-x) + log cos^2(x) - log(x^2+x+1)

log y = -x + 2log cos x - log(x^2+x+1)

1/y y' = -1 - 2tanx - (2x+1)/(x^2+x+1)

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

Now, you can massage that for a few more steps, to get something that pleases you

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