# discuss about the differentiability of $g(x)=|f(x)|$, where $f$ is a differentiable function

I want to discuss about the differentiability of $$g(x)=|f(x)|$$, where $$f$$ is a differentiable function

Example 1

Take $$f(x)=|x|$$, function is clearly not differentiable at $$x=0$$. Example 2

Take $$f(x)=|\sin(x)|$$, function is clearly not differentiable at the point $$x=n\pi$$ After taking few more examples like $$|x-1|, |\cos(x)|$$, it always seems to be the case that $$|f(x)|$$ is not differentiable at the points where $$f(x)=0$$

Observation: One thing is common in all the examples that some portion of $$f(x)$$ lies below $$x$$ axis. So I took another example

$$f(x)=x^2$$ but $$|f(x)|$$ is differentiable at the point where $$f(x)=0$$

Question 1:

Am I right in concluding that we can not just say in general setting that $$|f(x)|$$ is not differentiable at the points where $$f(x)=0$$?

When can we(I mean under what conditions can we )conclude that $$|f(x)|$$ is differentiable at points where $$f(x)=0$$. My hypothesis is that graph of $$f$$ should lie below $$x$$ axis.

Question 2:

Let $$f(x)$$ and $$g(x)$$ be two differentiable function, when can we conclude that $$|f(x)|+|g(x)|$$ is not differentiable at the points where $$f(x)=0$$ and $$g(x)=0$$

Example $$|sin(2-x)|+ |cos(x)|$$ are not differentiable at $$x=2+2\pi, x=(2n+1)\frac{k}{2}$$

Edits As mentioned by @Torsten Schoeneberg in comments, my hypothesis fails!! Grand Edit:

$$f(x)=|x|$$ then $$f'(x)=\text{sign}{(x)}$$

So let $$f(x)$$ be a differentiable function, and let $$g(x)=|f(x)|$$, then $$g'(x) =\text{sign}(f(x))f'(x)$$ Note that $$\text{sign}{(f(x))}=\begin{cases}{ -1 \quad \text{if } f(x)<0\\+1 \quad \text{if } f(x)>0 } \\{ 0 \quad \text{if } f(x)=0} \end{cases}$$

Am I going in right direction?

• To your hypothesis in question 1, check $f(x)= x^3$. Sep 19, 2018 at 23:00
• Yes to the first question: if $f$ is non-negative and differentiable, then $|f| \equiv f$ and thus the former is differentiable (in particular at its roots). There are plenty of such examples. Sep 19, 2018 at 23:03
• Perhaps check this conjecture: If $f$ is 0 at point $x=a$, and $f'(a)\neq 0$, then $|f|$ is not differentiable at $a$. Sep 19, 2018 at 23:08
• +1 for showing so much thought. Sep 19, 2018 at 23:08
• Concerning the "grand edit": Certainly that is one way to go for those $x$ where $f(x)\neq 0$, but be careful, we do not have that the derivative of the absolute value function at $0$ is $sign(0) =0$; rather, it is undefined. (The absolute value function is not differentiable at $x=0$, as you yourself noticed at the beginning of your post.) Sep 20, 2018 at 3:41

Hint: Try to show that if $f(x_0)\ne 0,$ then $f'(x_0)$ exists iff $|f|'(x_0)$ exists. And if $f(x_0)= 0$ and $f'(x_0)$ exists, then $|f|'(x_0)$ exists iff $f'(x_0)=0.$