is $$f(x) =\begin{cases} x^2\sin(\frac{1}{x}) \mbox{ for } x\neq 0 \\ 0 \mbox{ for } x= 0\end{cases}$$ a continuous function specially at point x=0?
And why being derivable its derivative is not continuous at x=0 since
$$f`(x) =\begin{cases} 2x\sin\frac{1}{x}-\cos\frac{1}{x} \mbox{ for } x\neq 0 \\ 0 \mbox{ for } x= 0\end{cases}$$?
I suspect this as to do with being derivable a function f $\mathbb{R}^n\rightarrow\mathbb{R}^m$ that don't imply its partial derivatives are continuous .this I don't get it so far.however partials derivatives if continuous is derivable.
Can you provide me a more general case than just one variable where f is derivable but its partials are not continuous?
$$0\leq \left|x^2\sin\left(\frac{1}{x}\right)\right|\leq x^2\to 0$$ if $x\to 0$. Then your function is continuous in $x=0$. Moreover, your function is derivable in $x=0$. Indeed,
$$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0} x\sin\left(\frac{1}{x}\right)=0.$$
The fact that $f'$ is not continuous on $x=0$ implies just that f is not $\mathcal C^1(\mathbb R)$ not that $f$ is not derivable on $\mathbb R$. This is the perfect example of a function derivable with a derivate not continuous.