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Find equation of tangent line

  1. Feb 10, 2010 #1
    Find the equation of the tangent line to the curve y=(lnx)^cosx at the point (pi/2, 1)?

    3. The attempt at a solution
    lny = cosx(ln(ln(x))) d/dx
    = -sinx(ln(ln(x))) + cosx/(ln(x)(x))
    y' = y(-sinx(ln(ln(x))) + cosx/(ln(x)(x)))
    this is the part where I get stuck
     
  2. jcsd
  3. Feb 10, 2010 #2

    jhae2.718

    User Avatar
    Gold Member

    When you get to
    [itex]y'=y\left[ {\frac{\cos\left(x\right)}{x\ln\left(x\right)}-\sin\left(x\right)\ln\left(\ln\left(x\right)\right)}\right][/itex]
    substitute [itex]\ln\left(x\right)^{\cos\left(x\right)}[/itex] for y.

    Then take [itex]y'\left(\frac{\pi}{2}\right)[/itex] as the slope of your tangent line and find what y-intercept will put the line through [itex]\left(\frac{\pi}{2},1\right)[/itex].
     
    Last edited: Feb 10, 2010
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