- #1

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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!

- Thread starter vikcool812
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- #1

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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

arildno

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Use the chain rule of differentiation.

- #3

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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

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[tex]\frac{d}{dx}\ln(|x|) = \frac{1}{x}[/tex]

is

- #5

- 537

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[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

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Thank You ! Everyone for your valuable suggestions .

- #7

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Hi g_edgar

THankyou for your hint , and my work was in the real numbers.

Best Regards

Riad Zaidan

THankyou for your hint , and my work was in the real numbers.

Best Regards

Riad Zaidan

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