D/dx of ln(|x|)

  • Thread starter vikcool812
  • Start date
  • #1
13
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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!
 

Answers and Replies

  • #2
arildno
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Use the chain rule of differentiation.
 
  • #3
13
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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
 
  • #4
607
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BEWARE... The formula
[tex]\frac{d}{dx}\ln(|x|) = \frac{1}{x}[/tex]
is wrong for complex [itex]x[/itex]
 
  • #5
537
3
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.
 
  • #6
13
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Thank You ! Everyone for your valuable suggestions .
 
  • #7
13
0
Hi g_edgar
THankyou for your hint , and my work was in the real numbers.
Best Regards
Riad Zaidan
 

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