Show that $f(x) = x\sin(1/x)$ is Differentiable everywhere where $x\ne0$. I have to show that  $f(x)= x\sin(1/x)$ is continuous everywhere differentiable everywhere where $x\ne 0$.

I can show the continuous property, and how it is not differentiable when $x=0$, but how would I go to prove that it is differentiable for all $x$, such that $x\ne 0$. 

Trying to put it into the definition of the derivative and simplifying does not work (I could not get a proper answer.)

Any hints on how to proceed would be greatly appreciated.
do you know that $\sin$, $x$ and $1/x$ are continuously differentiable everywhere (but $0$ in the last case)? if so just use the fact that a composition and a product of $C^1$ functions is $C^1$

edit: to be clear - $C^1$ means continuously differentiable
$$\text{Suppose the limit did exist then use the below to get a contradiction.}$$

$$\text{Limit Rule:} \ \lim x_n \to X \ \text{and} \ \lim y_n \to Y \Rightarrow \lim x_n y_n = XY$$