Derivatives of general absolute value functions

Question

Let $f : (a, b) \to \mathbb{R}$ be differentiable at $c \in (a, b)$, and $f(c) \neq 0$.

How do I show that $|f|'(c) = f'(c)$ if $f(c)>0$?

I know that $|f|'(c) = $ $\frac{|f(x)| - |f(c)|}{x-c}$.

Thanks in advance!

Answer 1

Case 1: $f(c)>0.$ then there is a neighborhood $U$ of $c$ with $U \subset (a,b)$ and $f(x)>0$ for all $x \in U.$

Then $|f|(x)=f(x)$ for all $x \in U.$ Thus $|f|'(c)=f'(c).$

Case 2: $f(c)<0.$ then there is a neighborhood $U$ of $c$ with $U \subset (a,b)$ and $f(x)<0$ for all $x \in U.$

Then $|f|(x)=-f(x)$ for all $x \in U.$ Thus $|f|'(c)=-f'(c).$