Answer
FALSE
Work Step by Step
Consider $f:R \to R $ such that $f(x)=|x^{2}+x|$
Since, $x^{2}+x\gt 0$ for all $x∈(-\infty,-1)∪(0,+\infty)$
we have that $f(x)=|x^{2}+x|=x^{2}+x$ for all $x∈(-\infty,-1)∪(0,+\infty)$
Hence, $f$ is differniable on $(-\infty,-1)∪(0,+\infty)$ and $f'(x)=2x+1$ for all $x∈(-\infty,-1)∪(0,+\infty)$
Take $x=-3$
$f'(-3)=2(-3)+1=-6+1=-5\ne 5=|2(-3)+1|$
Hence, the given statement is false.
