Multiple choice exercise on $f(x)= \frac {\sin x}{|x|+ \cos x}$ 
Let $f : \Bbb R \to \Bbb R$ be the function defined by $f(x)= \frac {\sin x}{|x|+ \cos x}$. Then
A.$f$ is differentiable at all $x \in \Bbb R$.
B.$f$ is not differentiable at $x =0$.
C.$f$ is differentiable at $x=0$ but $f'$ is not continuous at $x=0$.
D.$f$ is not differentiable at $x=\frac {\pi}{2}$.

Using the standard definition of derivative, I get that $f$ is differentiable at $0$ as well as $\frac {\pi}2$. This is as follows,
$\lim_{h \to 0} \frac {f(0+h)-f(0)}{h}=1=\lim_{h \to 0} \frac {f(0-h)-f(0)}{-h}$
and
$\lim_{h\to 0} \frac {f(\frac {\pi}2 +h)-f(\frac {\pi}{2})}{h}=0=\lim_{h \to 0} \frac {f(\frac {\pi}2 -h)-f(\frac {\pi}2)}{-h}$ using L'Hospital rule at one stage.
Hence I eliminate options B and D.
Since $0$ was the only doubtful point to check about differentiability, I choose option A as the answer.
I regret a little about not knowing how to check validity of option C. Can you tell me how to prove it wrong?
 A: A is true answer
By definition of the derivative
$$f'(0)=\lim_{h\to0}\frac{\frac{\sin h}{|h|+\cos h} - \frac{\sin 0}{0+1} }h=
\lim_{h\to0}\frac{\sin h}{h(|h|+\cos h)}
$$
Note that (using L'Hospital and then substitution)
$$
\lim_{h\to0^+}\frac{\sin h}{h(|h|+\cos h)}=\lim_{h\to0^+}\frac{\sin h}{h(h+\cos h)}=\lim_{h\to0^+}\frac{\cosh}{2h+\cos h - h\sinh)}=1
$$
Verify by yourself that also
$$
\lim_{h\to0^-}\frac{\sin h}{h(|h|+\cos h)}=
\lim_{h\to0^-}\frac{\sin h}{h(-h+\cos h)}=1
$$
Conclude that $f'(0)=1$

So you got that $f'(0)$ is defined, to show $f'$ is continuous at $x=0$ one need to show for example that 
$$\lim_{x\to0} f'(x)=f'(0)$$
$$\lim_{x\to0} f'(x) = 
\lim_{x\to0}\lim_{h\to0}\frac{\frac{\sin x+h}{|x+h|+\cos (x+h)} - \frac{\sin x}{|x|+\cos x} }h=
\lim_{h\to0}\lim_{x\to0}\frac{\frac{\sin x+h}{|x+h|+\cos (x+h)} - \frac{\sin x}{|x|+\cos x} }h=f'(0)
$$
The switching the order of limits is allowed when the function is uniformly continuous. A continuous function on a compact set is uniformly continuous, so everything is fine here, but I guess that you won't like this sort of answer:)

Using  $x\ne0$: $|x|'=\mathrm{sgn}(x)$ where
$$\mathrm{sgn}(x) = \begin{cases}1&x>0\\0&x=0\\-1&x<0\end{cases}$$
$$f'(x)=\frac{\cos (x)}{\left| x\right| +\cos (x)}-\frac{\sin (x) (\mathrm{sgn}(x)-\sin (x))}{(\left| x\right| +\cos (x))^2},\quad x\ne0$$
$$\lim_{x\to0^+} f'(x) = \frac{1}{0 +1}-\frac{0 (0-0)}{(0 +1)^2}=1$$
Similarly 
$$\lim_{x\to0^-} f'(x) = 1$$
Conclude 
$$\lim_{x\to0} f'(x)=f'(0)$$
and you done.
A: I feel you are looking for a descriptive answer, still leaving it here as an answer since it explains your doubt. 
From graph:

Where green is $f$, blue is $f'$, orange is $f''$. $f'$ is clearly continuous but not differentiable at $0$
A: As per the discussion with @Quintic in the above comments,
$f'(x)=\frac {1+|x|\cos x-\frac {x}{|x|} \sin x}{(|x|+\cos x)^2}$.
Now, $\lim_{h \to 0} f'(0+h)=\lim_{h \to 0} \frac {1+|h|\cos h-\frac {h}{|h|}\sin h}{(|h|+\cos h)^2}=\lim_{h \to 0}\frac {1+h\cos h-\sin h}{(h+\cos h)^2}$. (Where $h \gt 0$)
$\Rightarrow \lim_{h\to 0} f'(0+h)=\frac {1+0-0}{(0+1)^2}=1.$
Similarly, $\lim_{h\to 0} f'(0-h)=\lim_{h\to 0} \frac {1+h\cos h-\sin h}{(h+\cos h)^2}$ (Where $h \gt 0)$
$\Rightarrow \lim_{h\to 0} f'(0-h)=1$.
Hence $\lim_{h\to 0} f'(0+h)=\lim_{h\to 0} f'(0-h) \Rightarrow$ $f'$ is continuous at $x=0$.
