I.e., if |f| is differentiable at a point $a$, is $f$ differentiable at $a$ too? Obviously this will not work for a piecewise function $f$ whose absolute value is continuous, but I can't figure out the sufficient conditions for $f$ to be differentiable. It seems like continuity of $f$ at $a$ should force it to be differentiable at $a$, but I don't see any algebraic manipulations in $\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$ to make use of the continuity limit, that $\lim_{h\to 0}f(a+h)=f(a)$.
The differentiability of absolute value of a function implies that the function is differentiable
analysisderivatives
Best Answer
Suppose $|f|$ is differentiable at $a$ and $f$ is continuous at $a$.
There are two possibilities:
$1$: $f(a)\neq0$ Since $f$ is continuous, there exists some neighborhood around $a$ such that $f(a)\neq0$. That means that on this interval $U$, $f=|f|$ or $f=-|f|$. Either way, differentiability of $|f|$ implies differentiability of $f$.
$2$: $f(a)=0$. Note that $|f|'(a)=0$, so that tell us that $$\lim\limits_{h\to0}\frac{|f(a+h)|}h=0$$But this implies that $\lim\limits_{h\to0}\frac{f(a+h)}h$ does exist and also is $0$, since regardless of sign, $f(a+h)$ shrinks faster than $h$. This proves the desired statement.