Is $b|x|(\sin|x^2+x|)$ differentiable? $b$ can have any real value So I get that if only $\sin|x^2+x|$ was given it is not differentiable at $x=0$, but why does it become differentiable at $0$ when a factor of $b|x|$ is introduced? And if it does, then is the statement "A non differentiable function times a factor which becomes zero (at those non differentiable points) makes the whole function differetiable" universal?
For example, $y= |x^2-1|\sin\pi x$ is differentiable at both $1$ and $-1$ while the first factor, taken alone, is not. Is there a better logic?
 A: If $f$ is any function that satisfies $|f(x)| \le C x^2$ for $x$ close to $0,$ then $f'(0)=0.$ 
A: I will update this answer with more details when I have time, but here's something to think about:
Plug in $\sin |x^2+x|$ into the definition of the derivative, $\lim_{h \to 0} \dfrac {f(x+h)-f(x)}{h}$. Look at the limits as $h \to 0^{+}$, and as $h \to 0^{-}$. You are interested in it only when $x=0$, so you can plug that in for $x$.
Now do the same with $b |x| \sin |x^2+x|$ and compare the results. (also note the trivial case when $b=0$, since you let $b \in \mathbb{R}$ )
A: If $b=0$, then the function is obviously differentiable. If $b\ne0$, differentiability of $f$ does not depend on $b$, so we can assume $b=1$.
The derivative can be computed by using that the derivative of $|x|$ is $\operatorname{sgn} x$, for $x\ne0$. Thus we see there are problems at $0$ and $-1$.
On the interval $(-1/2,1/2)$ we have
$$
f(x)=x\sin(x^2+x)
$$
so the function is differentiable at $0$.
Look at what happens in the interval $(-2,-1)$ and in the interval $(-1,0)$ and conclude.
