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Logarithmic Differentiation

  1. Mar 4, 2009 #1
    y = (sin2x)3(x4-4x)6 divided by (cosx) + e3x

    i came out with an answer

    y' = (sin2x)3(x4-4x)6divided by (cosx) + e3x [3cot2x + 24x3-24 divided by x4-4x + Tanx + 3x

    could someone tell me if im right?
    i dont know if this is what im supose to do,
    if you want me to write out exactly what i did i can.
  2. jcsd
  3. Mar 4, 2009 #2


    User Avatar
    Science Advisor

    You titled this "logarithmic differentiation" but I see no logarithm.
  4. Mar 4, 2009 #3
    i'm supose to solve it using logartimic differentiation, so when i do it out i get

    lny = 3ln(sin2x) + 6ln(x^4-4x) - ln(cosx) + 3xln(e)
  5. Mar 4, 2009 #4


    Staff: Mentor

    Your last two terms came from cos(x) + e^(3x) in the denominator. ln(a + b) [itex]\neq[/itex] ln a + ln b
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