Calculus Help

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use logarithmic diff. to find the derivative of the function. Show steps please! so I can see how it is done. Thank you so much!

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

  • Calculus Help -

    y = (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 -

    log 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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