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D/dx of ln(|x|)

  1. Oct 14, 2009 #1
    I know d/dx of ln(x) is 1/x , x>0.
    A website says d/dx of ln(|x|) is also 1/x for x not = 0 .
    is that true , i am unable to prove it!
  2. jcsd
  3. Oct 14, 2009 #2


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    Use the chain rule of differentiation.
  4. Oct 14, 2009 #3
    Hi vikcool812 :
    Note that ln(|x|)=ln(x) when x>0 and
    =ln(-x) when x<0
    So d/dx (ln(|x|))=d/dx(ln(x)) when x>0 and
    d/dx (ln(|x|))=d/dx(ln(-x)) when x<0 therefore
    d/dx(ln(x)) =1/x when x>0 by definition of ln(x) and
    d/dx(ln(-x))=-1/-x = 1/x when x<0 by chain rule so in both cases we have
    d/dx (ln(|x|))=1/x for not x=0
    Best Regards
    Riad Zaidan
  5. Oct 14, 2009 #4
    BEWARE... The formula
    [tex]\frac{d}{dx}\ln(|x|) = \frac{1}{x}[/tex]
    is wrong for complex [itex]x[/itex]
  6. Oct 14, 2009 #5
    Remember that when dealing with absolute values, with either derivatives or limits, it is best to directly use the definition of the absolute value. I.e.,
    [tex]|x| = \begin{cases} x &, x\geq 0 \\ -x &, x\leq 0 \end{cases}[/tex]
    This is of course what rzaidan used in his solution.
  7. Oct 14, 2009 #6
    Thank You ! Everyone for your valuable suggestions .
  8. Oct 15, 2009 #7
    Hi g_edgar
    THankyou for your hint , and my work was in the real numbers.
    Best Regards
    Riad Zaidan
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