If $f$ is a derivative then is $|f|$ also a derivative? If $f$ is a derivative then,  is $|f|$ also a derivative?
If $f$ is Riemann integrable then it's true. But, if it's not the case,then is it true?
 A: This is not true even if $f$ is integrable.
Consider $f(x) = \begin{cases}\cos(x^{-1}), & 0 < x \leqslant   1 \\ 0, & x= 0 \end{cases}$ and $g(x) = |f|(x) = \begin{cases}|\cos(x^{-1})|, & 0 < x \leqslant   1 \\ 0, & x= 0 \end{cases}$ 
We can show that $f = F'$ where $F(x) = \int_0^x f(t) \, dt$. This is easy to show for $0 < x \leqslant 1$ where $f$ is continuous. It is also true but harder to show that $F'(0) = f(0) = 0$.
Now if $g = G'$ on $[0,1]$ and $g$ is integrable then $\int_0^x g(t) \, dt = G(x)$ by the FTC with $G'(0) = g(0)$.
However, 
$$G'(0) = \lim_{h \to 0} \frac{1}{h}\int_0^h g(t) \, dt = \lim_{h \to 0} \frac{1}{h}\int_0^h |\cos(t^{-1})| \, dt = \frac{2}{\pi} \neq g(0)$$
See here for a proof that $G'(0) = 2/\pi$.
Addendum
To prove that $F'(0) = 0$, note that $F(0) = 0$ and with a variable change $y = t^{-1}$,
$$F(x) = \int_0^x \cos (t^{-1}) \, dt = \int_{x^{-1}}^\infty \frac{\cos y}{y^2} \, dy$$
Integrating by parts we get,
$$F(x) = \frac{\sin(x^{-1})}{x^{-2}} + 2 \int_{x^{-1}}^\infty \frac{\sin y}{y^3} \, dy$$
Thus,
$$0 \leqslant \left|\frac{F(x)-F(0)}{x} \right| \leqslant \frac{|\sin(x^{-1})|}{x^{-1}} + \frac{2}{x} \int_{x^{-1}}^\infty \frac{|\sin y|}{y^3} \, dy \\ \leqslant x + \frac{2}{x} \int_{x^{-1}}^\infty \frac{1}{y^3} \, dy = 2x$$
By the squeeze theorem it follows that
$$F'(0) = \lim_{x \searrow 0}\frac{F(x)-F(0)}{x} = 0 $$
A: Consider $f(x)=x$ which is differentiable everywhere but $|f(x)|$ is not differentiable at zero.
