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Find the derivative using Logaritmic Differentiation

  1. Mar 5, 2009 #1
    y = (sinx)2x

    LNy = 2xLN(sinx) + (1 over sinx)(cosx)(2x)


    y' = (sinx)2x [2cosx over sinx + 2xcotx]


    y = (cosx)cosx

    i did it the same way as above
    Answer i got was

    y' = (cosx)cosx [ (-sinx)(cosx) + sinx]

    am i anywhere right?
  2. jcsd
  3. Mar 5, 2009 #2
    How comes the right term??

    [tex]ln(y) = 2xln(sinx)[/tex]
    [tex]~~\frac{y'}{y} = 2(ln(sinx)+xcotx)[/tex]
    [tex]~~y' = 2sinx^2x(ln(sinx)+xcotx)[/tex]

    Nope. Please try again. There should be a ln() in your answer.
  4. Mar 5, 2009 #3
    for the second one , would this be right?

    y' = (cosx)cosx[(-sinx)(LN(sinx) - sinx]
  5. Mar 5, 2009 #4


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    Yes, it would.
  6. Mar 6, 2009 #5
    I don't think so. I think it should be ln(cosx) instead of ln(sinx). Typo?
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