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Math Help - Differentiate

  1. #1
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    Differentiate

    Differentiate y=\frac{lnx}{ln(x-1)} with respects to x

    y=\frac{lnx}{ln(x-1)}= \frac{(ln(x-1))(\frac{1}{x})-lnx(\frac{1}{x-1})}{[ln(x-1)]^2}
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Punch View Post
    Differentiate y=\frac{lnx}{ln(x-1)} with respects to x

    y=\frac{lnx}{ln(x-1)}= \frac{(ln(x-1))(\frac{1}{x})-lnx(\frac{1}{x-1})}{[ln(x-1)]^2}
    it is correct but make it well order
    \frac{dy}{dx}=\left(\frac{lnx}{ln(x-1)}\right)^{'}= \frac{(ln(x-1))(\frac{1}{x})-lnx(\frac{1}{x-1})}{[ln(x-1)]^2}
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  3. #3
    Senior Member BAdhi's Avatar
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    also it has to be \ln{|x-1|} not \ln{(x-1)}
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  4. #4
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    Quote Originally Posted by BAdhi View Post
    also it has to be \ln{|x-1|} not \ln{(x-1)}
    No it doesn't. Only if you are arriving at a logarithm through integration do you need the modulus signs.
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  5. #5
    Senior Member BAdhi's Avatar
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    please elaborate

    Only if you are arriving at a logarithm through integration do you need the modulus signs.
    Prove It, can you please explain it more because I quite don't get it
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  6. #6
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    \displaystyle \ln{(x)} is a continuous function on \displaystyle (0, \infty) and is differentiable on \displaystyle (0, \infty). Its derivative is \displaystyle \frac{1}{x}.

    But \displaystyle \frac{1}{x} is defined on all \displaystyle x \in \mathbf{R}\backslash \{0\} , so should have an antiderivative on \displaystyle (-\infty, 0) as well.

    So we need to think outside the box a little. Remember that \displaystyle |x| = \begin{cases} x \textrm{if }x \geq 0\\ -x \textrm{if }x < 0\end{cases}.

    What is the derivative of \displaystyle \ln{|x|}?
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  7. #7
    Senior Member BAdhi's Avatar
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    It should be \frac{1}{|x|}.

    Ok, let me get this straight. When you mention \ln{(x)}, variable x's domain should always be (0,\infty). And the derivative of it then will become \frac{1}{x}. For a variable y \in (-\infty,\infty) it should be \ln{|y|} and the derivative is \frac{1}{|y|} (we can verify it by knowing the slope of the curve \ln{(x)} is always positive).

    Have I got it right of worse?
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  8. #8
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    No, the derivative is not \displaystyle \frac{1}{|x|}.

    Remember that \displaystyle \ln{|x|} = \ln{(x)} if \displaystyle x > 0 and \displaystyle \ln{|x|} = \ln{(-x)} if \displaystyle x < 0.

    So to evaluate the derivative of \displaystyle \ln{|x|}, you need to differentiate each of those two cases. You should notice something...
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  9. #9
    Senior Member BAdhi's Avatar
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    sorry sir, forgot that the curve of the \ln{|x|} is \ln{(x)} mirrored around the y axis. So derivative should still be \frac{1}{x}

    thanks.
    Last edited by BAdhi; June 12th 2011 at 07:10 AM. Reason: added some text
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  10. #10
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    Correct, so the antiderivative of \displaystyle \frac{1}{x} is \displaystyle \ln{|x|}
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