Is d/dx of ln(|x|) = 1/x? Proving the Claim

  • Context: Undergrad 
  • Thread starter Thread starter vikcool812
  • Start date Start date
Click For Summary
SUMMARY

The derivative of ln(|x|) is confirmed to be 1/x for all x ≠ 0. This is established by applying the chain rule of differentiation, where ln(|x|) is expressed as ln(x) for x > 0 and ln(-x) for x < 0. The calculations show that d/dx(ln(x)) = 1/x for x > 0 and d/dx(ln(-x)) = 1/x for x < 0, thus proving the claim. Caution is advised when extending this formula to complex numbers, as it does not hold true in that context.

PREREQUISITES
  • Understanding of basic calculus concepts, specifically differentiation
  • Familiarity with the natural logarithm function, ln(x)
  • Knowledge of the chain rule in differentiation
  • Concept of absolute values in mathematical expressions
NEXT STEPS
  • Study the chain rule of differentiation in depth
  • Explore the properties of the natural logarithm function, ln(x)
  • Learn about the implications of absolute values in calculus
  • Investigate the behavior of derivatives in complex analysis
USEFUL FOR

Students of calculus, mathematics educators, and anyone interested in understanding the nuances of logarithmic differentiation and its applications in real and complex analysis.

vikcool812
Messages
13
Reaction score
0
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!
 
Physics news on Phys.org
Use the chain rule of differentiation.
 
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
 
BEWARE... The formula
\frac{d}{dx}\ln(|x|) = \frac{1}{x}
is wrong for complex x
 
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.,
|x| = \begin{cases} x &amp;, x\geq 0 \\ -x &amp;, x\leq 0 \end{cases}
This is of course what rzaidan used in his solution.
 
Thank You ! Everyone for your valuable suggestions .
 
Hi g_edgar
THankyou for your hint , and my work was in the real numbers.
Best Regards
Riad Zaidan
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K