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Is $|\ln|x||$ differentiable for all $x$ is defined and continuous?

I can see that on the graph that it is not differentiable at $-1$ and $1$, but how can I prove it?

So I look at $\lim_{h\to 0+} \frac{|\ln(1-h)|-|\ln(1)|}{h}=\lim_{h\to 0+} \frac{|\ln(1-h)|}{h}$ because it is positive $(0^{+})$ we can say $\lim_{h\to 0+} \frac{\ln(1-h)}{h}$ applying L'Hôpital $\lim_{h\to 0+} \frac{\frac{-1}{1-h}}{1}=-1$?

Lonidard
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gbox
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5 Answers5

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At points where $f(x)\ne0$ there is no problem in differentiating $|f|$, because the function is positive/negative in a neighborhood of the point.

So all you need is to check where $f(x)=0$. In this case at $1$ and $-1$. However, since the function is clearly even, we can just look at $1$: $$ \lim_{x\to1^+}\frac{|\log|x|\,|-|\log|1|\,|}{x-1}= \lim_{x\to1^+}\frac{\log x}{x-1}=1 $$ whereas $$ \lim_{x\to1^-}\frac{|\log|x|\,|-|\log|1|\,|}{x-1}= \lim_{x\to1^-}\frac{-\log x}{x-1}=-1 $$ So the function is not differentiable at $1$ and $-1$, but it is everywhere else (provided it is defined to begin with).

Yes, your attempt is good.

egreg
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  • can we say that graph exhibits cusp at $x=\pm 1$ – Ekaveera Gouribhatla Dec 07 '15 at 10:42
  • @EkaveeraKumarSharma Not really a cusp. It's a jump (of 1st order discontinuity) for the derivative, but this does not change the fact that the derivative does not exist for $x=\pm1$. – yo' Dec 07 '15 at 10:44
  • I think the limit calculation I did is wrong, I get the same result for $1^{+} $ and $ 1^{-}$ – gbox Dec 07 '15 at 10:48
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    @EkaveeraKumarSharma In Italian it's called “punto angoloso” (angle point, the direct English translation). – egreg Dec 07 '15 at 12:36
  • @gbox You have surely an error, because the numerator is positive by definition, while the denominator is negative for the limit from the left and positive for the limit from the right. – egreg Dec 07 '15 at 12:36
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Assuming $f(x) : \mathbb{R}\setminus \{0\} \to \mathbb{R}$.

We can rewrite $f(x) = \left|\ln\left|x\right|\right|$ as:

$$ f(x) = \begin{cases} \ln(x), & x \geq 1 \\ -\ln(x), & 1 \geq x > 0 \\ -\ln(-x), & 0 > x \geq -1 \\ \ln(-x), & -1 \geq x \end{cases} $$

We see that on each of the intervals $(-\infty, -1), (-1, 0), (0, 1), (1, \infty)$, $f(x)$ is both continuous and differentiable, due to the continuity and differentiability of $\ln(x)$ on $(0, 1)$.

We thus have 2 possible $x$ values where $f$ is not differentiable, $-1$ and $1$. Note that we don't have to consider $x=0$ as $f$ is not defined there.

Lets look at $x = 1$.

For $x \geq 1$ we have $f'_+(x) = \frac{1}{x}$, and for $1 \geq x > 0$ we have $f'_{\vphantom+-} (x) = -\frac{1}{x}$. Thus:

$$\lim_{x \to 1+} f'_+(x) = 1 \neq \lim_{x \to 1-} f'_{\vphantom+-}(x) = -1$$

Tormod Haugland
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The facts:

  1. $|x|$ is not differentiable at $x = 0$
  2. The derivative of $\ln |x|$ vanishes no-where on $\mathbb{R}$

implies that $ |\ln |x| |$ is not differentiable at $x$ for which $\ln |x| = 0$ i.e. $x= \pm 1$

  • is there a way to turn a function that is not differentiable to be differentiable?

    Or the way you solve it, I can use everywhere?

     – gbox Dec 07 '15 at 10:22
  • @gbox No, you can not use it when the outer function has zero derivative at zero. – yo' Dec 07 '15 at 10:25
  • @yo' This method says nothing about $|x|^2$ or any other powers of $|x|$ –  Dec 07 '15 at 10:29
  • @yo' strictly increasing functions –  Dec 07 '15 at 10:31
  • I should not have used the term well behaved. Sorry about that. –  Dec 07 '15 at 10:32
  • @yo' oops. strictly increasing and non-zero at x=0. I understood why this method is wrong. But I am trying to find the class of functions for which it is valid –  Dec 07 '15 at 10:34
  • @yo' yes, you are right. $f(x)$ must have non-vanishing derivative. –  Dec 07 '15 at 10:52
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Assuming a function from reals to reals:

$$\frac{\text{d}}{\text{d}x}\left(\left|\ln|x|\right|\right)=\frac{\ln|x|\left(\frac{\text{d}}{\text{d}x}\left(\ln|x|\right)\right)}{|\ln|x||}=\frac{\ln|x|\cdot\frac{\frac{\text{d}}{\text{d}x}\left(|x|\right)}{|x|}}{|\ln|x||}=\frac{\ln|x|\cdot\frac{x}{|x|}}{|\ln|x||}=\frac{\ln|x|}{x|\ln|x||}$$

Jan Eerland
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  • And this is supposed to solve what exactly? It looks to me like blind and foolish application of the derivative without seeing any details and consequences. – yo' Dec 07 '15 at 10:23
  • I gave him the derivative, now the OP can take the effort to see the consequences – Jan Eerland Dec 07 '15 at 10:25
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The first step would be to differentiate it

Check out the derivative here (hint: $|x| = (x^2)^{0.5}$ for reals)

$\frac{d}{dx} |ln|x|| = \frac{ln|x|}{x|ln|x||}$

$lim_{x -> 0^+} \frac{ln|x|}{x|ln|x||} = + \infty$

while

$lim_{x -> 0^-} \frac{ln|x|}{x|ln|x||} = - \infty$

So the function IS differentiable, but not at certain points.

Note that the derivative contains the logarithm function which is not defined for zero. To understand why this is not defined at zero, you just need to know what a logarithm is.

$x = ln(y)$ is the inverse of $y = e^x$

there isn't any number $x$ that will produce zero for $e^x$

so $ln(0)$ is not defined

also the function doesn't make any sense at $+1$ and $-1$, just plug those values into the derivative function and see what happens.

sav
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