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Am I right?

  1. Dec 3, 2005 #1
    Ok, I worked this derivative problem, but my book has a different answer than what I got. I'm not sure why.

    I need to compute the derivative of log base3 of (x^3 + 2x)

    I came out with:

    1/((x^3 + 2x)ln3)

    The book says that the answer is the same as mine except the numerator has a (3x^2 + 2) in it.

    Why would you put the derivative of the x value of the log function in the numerator.

    My book says that D log base a of x = 1/xlna

    help?
     
  2. jcsd
  3. Dec 3, 2005 #2

    LeonhardEuler

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    Gold Member

    The book is right. Don't forget the chain rule. It is true that
    [tex]\frac{d\ln{x}}{dx}=\frac{1}{x}[/tex]
    , but
    [tex]\frac{d\ln{u}}{dx}=(\frac{1}{x})(\frac{du}{dx})[/tex]
     
  4. Dec 3, 2005 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Have you heard of the "chain rule"?
    The derivative of log3{x^3 + 2x}[/sub] is
    [tex]\frac{1}{x^2+2x}(3x^2+ 2}[/tex].
     
  5. Dec 3, 2005 #4
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