At which points the given function $f$ is not differentiable? Let $f(x) = \cos (|x-5|) + \sin (|x-3|) + |x + 10 | ^ 3 - (|x| + 4 )^2$;
at which points the function $f$ is not differentiable?
Since $|x-a|$ is not differentiable at any real number $x= a$. so, the function $f$ is not differentiable at $x=5,3,-10$ and $0$.
Am I right? Please guide me.
 A: We first deal in detail  with the $|x+10|^3$ part. Looks like there might be trouble at $x=-10$. But there isn't! 
There are various ways to prove this. One way is to go back to the definition of the derivative. Let $p(x)=|x+10|^3$. We have
$$\frac{p(-10+h)-p(-10)}{h}= \frac{|h|^3}{h}=h^2\frac{|h|}{h}\tag{1}$$
As $h$ approaches $0$, the right-hand side of (1) approaches $0$, and quickly at that! So the derivative of $|x-10|^3$ exists at $x=-10$, and indeed is $0$.
You will find a similar phenomenon for the cosine part of your function. 
It is differentiable at $x=5$. Equivalently, let us see that $\cos| t|$ is differentiable at $0$.  
We could again go back to the definition of the derivative, but for a change, let's not. The cosine function is even, $\cos (-u)=\cos(u)=\cos(|u|)$, so our problem reduces to whether $\cos t$ is differentiable at $0$. It sure is!
The sine part, however, will give non-differentiability at $3$. We leave this part to you for a while. For the part about $(|x|+4)^2$, expand and you will see that we have non-differentiablity at $0$.
