continuous f(x) and continuous f '(x) relation IF $g(x)$ is prove to be not continuous at one point p. 
$g'(x)$ is also not continuous? 
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question is asking to show that g'(x) is not continuous at 0.

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let say we have 
$$g(x) =\begin{cases} 
x^2 sin(1/x) & x != 0 \\  
0, x = 0 \end{cases}$$
i'm going to show that  g(x) is not continuous
if  $|x-0| < \delta \\$
, and define the  x < $\delta$  and   $\delta  <= \epsilon/2$.
h = x + $\epsilon$
|g(h) - g(0) | = $|x^2sin(1/x) - 0|  $
then   if we put x as $\epsilon/2 $
$$\epsilon<(\epsilon/2)^2 sin( 2/\epsilon) $$ 
then
$$ |x^2sin(1/x) - 0|   > \epsilon$$
Therefore, This function is not continuous at point 0. 
can we conclude this to $g'(x) $ is not also continuous?
 A: Your function $g(x)$ is continuous everywhere. So you will not be able to show it is not continuous. 
We now look at a related question that you do not explicitly raise. 
If a function is differentiable at $p$, then it is automatically continuous at $p$. For if 
$$\lim_{x\to p}\frac{g(x)-g(p)}{x-p}=f'(p),$$ then 
$$\lim_{x\to p} (x-p)\frac{g(x)-g(p)}{x-p}=\lim_{x\to p}(x-p)f'(p)=0$$
and therefore $\lim_{x\to p}g(x)-g(p)=0$, meaning that $\lim_{x\to p}g(x)=g(p)$. This says $g(x)$ is continuous at $p$.
Equivalently, if $g(x)$ is not continuous at $p$, then $g(x)$ is not differentiable at $p$.  However, you will not be able to use this kind of argument to show that $g'(x)$ is not coninuous at $0$, since in fact $g(x)$ is continuous at $0$. 
Added: The question has changed. Directly from the definition of the derivative, you can prove that $g'(0)=0$. For we are looking at the limit of $\frac{x^2\sin(1/x)-0}{x-0}$, that is, of $x\sin(1/x)$. The $\sin(1/x)$ part stays bounded, and the $x$ part approaches $0$, so the product approaches $0$. 
To show $g'(x)$ is not continuous at $0$, calculate $g'(x)$ for $x\ne 0$ using the ordinary rules of differentiation. We get $-\cos(1/x)+2x\sin(1/x)$. It is not hard to show that the limit of this as $x\to 0$ does not exist. The $2x\sin(1/x)$ part behaves very nicely, but $-\cos(1/x)$ oscillates wildly as $x\to 0$. 
