Derivative of a "pointwise" function? What does it mean to take a derivative of a point-wisely defined function, in this case:
$\
 f(x) =
  \begin{cases}
   \frac{\sin x}{x}
 & \text{if } x ≠ 0 \\
   0       & \text{if }x=0
  \end{cases}
$  
 A: For each $x \neq 0$, $f'(x)$ is just the derivative of $\sin{(x)}/x$, since $$\lim \limits_{h \to 0} \dfrac{f(x + h) - f(x)}{h} = \lim \limits_{h \to 0} \dfrac{\dfrac{\sin{(x + h)}}{x + h} - \dfrac{\sin{(x)}}{x}}{h}.$$
But, you have to worry about if the derivative exists at $0$.  To be differentiable, it has to be continuous first, so you should check that $\lim \limits_{x \to 0} f(x) = f(0)$, i.e., $\lim \limits_{x \to 0} \dfrac{\sin{(x)}}{x} = 0$.
(Also, recall that $\lim \limits_{x \to 0} \dfrac{\sin{(x)}}{x} = 1$ which you can verify using L'Hopital's rule).
A: It means (I'm assuming your question is at $x = 0$) applying the definition of the derivative to the function in question. That definition is not "apply all the rules of differentiation", but rather, 
$$
f'(0) = \lim_{h \to 0} \frac{ f(0 + h) - f(0) }{h}. 
$$
If you write out that limit in this case, you get 
$$
f'(0) = \lim_{h \to 0} \frac{ \frac{\sin h}{h} - 0 }{h}. 
$$
Now you have to actually compute that limit (if it exists). 
"But why can't I just use the formulas?" you might ask. Well, look at the hypotheses behind each formula, and you may perhaps discover why. 
