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Math Help - Derivative of y=ln(lnx)

  1. #1
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    Derivative of y=ln(lnx)

    Find the equation of the tangent line to the curve y=ln(lnx) which is perpendicular to the line ex+y =0

    So I took the derivative and got 1/(xlnx)
    I know that the slope of the normal line is 1/e so i substituted that in for y' and tried solving for x. this is where I was stuck
    1/e = 1/(xlnx)
    xlnx = e
    Last edited by mr fantastic; April 30th 2010 at 03:37 PM. Reason: Restored deleted question.
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  2. #2
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    Quote Originally Posted by john-1 View Post
    Find the equation of the tangent line to the curve y=ln(lnx) which is perpendicular to the line ex+y =0

    So I took the derivative and got 1/(xlnx)
    I know that the slope of the normal line is 1/e so i substituted that in for y' and tried solving for x. this is where I was stuck
    1/e = 1/(xlnx)
    xlnx = e
    Looks correct to me.

    x \ln x = e \implies \ln x^x = e \implies x^x = e^e

    Therefore x=e.

    It's sort of a weird way to solve for x but it gives the correct solution.
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