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.
Let $g(x) = x$, $h(x) = \sin(x)$ and $k(x) = \frac{1}{x}$ defined for all $x \neq 0$. Then what you have is that $f(x) = g(x) \cdot h(k(x))$. In other words, $f$ is created by the product and composition of "nice" functions. So, if you know that $g, h,$ and $k$ are differentiable (when $x \neq 0$) and if you know the product rule and chain rule, then the differentiability of $f$ should follow.
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
$$\text{Suppose the limit did exist then use the below to get a contradiction.}$$
$$\text{Limit Rule:} \ \lim x_n \to X \ \text{and} \ \lim y_n \to Y \Rightarrow \lim x_n y_n = XY$$
