Determine where $k(x)=\left\vert\sin x\right\vert$ is differentiable and find the derivative. 
Determine where $k(x) = \left\vert\sin x\right\vert$ is differentiable and find the derivative.

Note that $f(0)=0$, then $|f(0)|=0$. We know that $k(x)$ is differentiable in $\mathbb{R}$, then $|k(x)|$ is differentiable in $\mathbb{R}$, but $f'(0)=0$, so $k(x)$ is differentiable in $\mathbb{R}$ \ $\{0\}$.
Is everything right?
 A: It is not true that $k$ differentiable in $\Bbb R$ implies $|k|$ differentiable in $\Bbb R \setminus \{0\}$, only that it is differentiable in $\Bbb R \setminus \{\text{zeroes of $k$}\}$, at least.
That being said, we have that $|\sin x|$ is differentiable at least in $\Bbb R \setminus \{ k\pi \mid k \in \Bbb Z \}$. By periodicity, we only have to check if $|\sin x|$ is differentiable at $0$ and $\pi$. The function is not differentiable at the these points, as you can see by looking at the lateral limits in the definition of derivative. So $|\sin x|$ is differentiable precisely in $\Bbb R \setminus \{ k\pi \mid k \in \Bbb Z \}$.
A: No, for a few reasons. For one, the claim that differentiability of $k(x)$ implies that of $\left\vert k(x) \right\vert$ is false: For example, the identity function $x \mapsto x$ is differentiable on $\Bbb R$, but the absolute value function $x \mapsto |x|$ is not differentiable at $0$, as the left- and right-hand limits of its difference quotient there do not agree.
A: Hint: $|f(x)|$ is non-differentiable at points where $f(x)=0$.
A: Use the concept behind chain rule: if $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$ then $g(f(x))$ is differentiable at $a$.
Here $f(x) = \sin x$ which is differentiable everywhere and $g(x) = |x|$ which is differentiable everywhere except $x = 0$. Hence we expect $g(f(x)) = |\sin x|$ to be differentiable everywhere except when $\sin x = 0$.
