Given $f(x) =\begin{cases} x\sin(1/x) & x \neq 0,\\ 0 & x= 0, \end{cases} $ why is $f'(0)$ not zero? When we use our standard product and chain rules for when $x$ not equal to $0,$ our derivative function works.
But when $x=0,$ $f'(x)$ does not exist. But why can't we just use standard differentiation rule on $x=0$? Won't that mean $f'(0) = 0$?
 A: It is unfortunate that math gets taught in such a way that people think it's done by applying rules of that sort. It relies on many such rules, but they are only a tool.
Actually understanding the subject as opposed to applying such rules will remind you that the value of the derivative at a point is the slope of the curve at that point, and that depends on what happens in some interval surrounding that point, not just on the value of the function at that point alone. Thus when you are told that if $f(x) = 5$ then $f'(x) =0,$ it means that if there is some interval such that for all values of $x$ in that interval $f(x) = 5,$ then $f'(x)=0$ for every value of $x$ in the interior of that interval.
To say that $f'(x)=0$ because $f(x) = 5,$ when $f(x) = 5$ at just an isolated point, is nonsense. For example, suppose $f(x) = x^2 - 4.$ Then $f(3) = 3^2-4=5.$  But that doesn't mean $f'(3)=0.$ In fact $f'(x) = 2x,$ so $f'(3)=6.$
Since the usual rules don't apply at $0$ for the function you give us, you need to go back to the definition:
$$
f'(0) = \lim_{h\to0} \frac{f(0+h) - f(0)} h.
$$
If that limit exists then $f'(0)$ exists; otherwise it doesn't.
A: Standard differentiation gives us:
$$f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x-0} = \lim_{x \to 0} \sin \left( \frac 1x \right)$$
It's well-known that the limit doesn't exist. Hence $f'(0)$ doesn't exist.
A: I think what you are missing is the fact that, while we usually speak of the derivative at a point, actually the derivative tells us something about the function in a neighbourhood of the point. Therefore you cannot just take the definition $f(0)=0$ and take the derivative of $0$, because there is no neighbourhood of $0$ where this is true.
To make this more obvious, define some function $g$ as follows:
$$g(x)=\begin{cases}x & x\ne 0\\0 & x = 0\end{cases}$$
Obviously this function is simply $g(x)=x$. However if we would use just the functional term given for the point in question and derive that, we'd get at the (wrong) result:
$$g'(x)=\begin{cases}1 & x\ne 0\\0 & x = 0\end{cases}$$
Of course the correct derivative is $g'(x)=1$ everywhere, as it doesn't matter how we write down that function.
A: That does not even exists.
In fact, $$f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x-0} = \lim_{x \to 0} \sin \left( \frac 1x \right)$$
Does not exist . Just consider: $$z_n  =\frac{1}{n\pi}\to 0~~~and~~~y_n  =\frac{1}{\frac{\pi}{2}+2n\pi}\to 0$$
But for all n we have, 
$$\sin \left( \frac{1}{z_n} \right) = 0~~~~and ~~~~\sin \left( \frac{1}{y_n} \right) = 1$$
A: Whenever you have a function with cases for $x=a$ and $x \neq a$, to calculate $f'(a)$ you have to use the definition of the derivative and do it with the limit.
In your case you have a function with cases for $a=0$. Therefore to determinate $f'(a)=f'(0)$ you have to go with the definiton.
$$f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x-0} = \lim_{x \to 0} \sin \left( \frac 1x \right)$$
Since that limit does NOT exist, you can conclude that $f'(0)$ doesn't exist.
A: Another example.  Let
$$
\phi(x) = \begin{cases}
x, & x \ne 0
\\ \; \\
0, &x = 0\end{cases}\tag{1}
$$
In fact, this is exactly the function $\phi(x) = x$, but of course we can write it as (1).
But the point is: you cannot compute $\phi'(0)$ by differentiating the $0$ in (1) to get $0$. That method is just wrong.
