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help!! differentiate wrt x: e^-cx^2*cos(logx)?

CALCULUS PEOPLE!

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  • 1 decade ago
    Favorite Answer

    Expression to differtiate

    e^-cx^2cos(log x)

    Let y= e^-cx^2cos(log x)

    y= e^{-cx^[2cos(log x)]}

    Take log on both sides to base e

    log y = log { e^{-cx^[2cos(log x)]}}

    = {-cx^[2cos(log x)]} * log e

    { because log a^ [exp] = [exp] * log a}

    log y = {-cx^[2cos(log x)]}

    (-1/c)*log y = x^[2cos(log x)]

    let d = -1/c constant

    d*log y = x^[2cos(log x)]

    Again take log on both sides to base e

    log (d *log y) = log {x^([2cos(log x)])}

    log d+log(log y) = 2cos(log x) *log x

    Now differentiate w.r.to x

    1/(log y) * (1/y) * dy/dx = 2cos(log x) * 1/x + log x * (-2sin(log x) * (1/x) )

    (1/(y*log y) *dy/dx = (1/x) { 2cos (log x) - 2 (log x) * sin(log x) }

    dy/dx = y*(log y) * (1/x) * { 2cos (log x) - 2 (log x) * sin(log x) }

    dy/dx = e^{-cx^[2cos(log x)]} * {-cx^[2cos(log x)]} * (1/x) * { 2cos (log x) - 2 (log x) * sin(log x) }

    I hope i remember maths i've not been in touch with maths for years

  • 1 decade ago

    yeah,u need to put in more paranthesis to clear whether -cx^2 is with -c or with e...??!!

  • 1 decade ago

    The question is ambiguous. Please edit the question by using more parentheses to make it clear.

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