differentiablility of f(x)=$x^m sin(1/x^n)$ 
my attempt:
i have to choose one from option A and option D. option B can be eliminated by taking m=1,n=2. option C can also be eliminated by taking m=4, n=3.  plz help from choosing one from A and C.
thanks.
 A: I'll assume $n\ge1$, otherwise the problem is trivial.
Use the definition of derivative: the derivative exists if and only if
$$
f'(0)=\lim_{x\to 0}x^{m-1}\sin\frac{1}{x^n}
$$
exists and is finite.
If $m>1$, this limit exists and is zero, because, for $x\ne0$,
$$
-|x^{m-1}|\le x^{m-1}\sin\frac{1}{x^n}\le |x^{m-1}|
$$
and the squeeze theorem applies, so $f'(0)=0$. If $m\le1$ the limit does not exist.
No hypothesis whatsoever is needed on $n$. So, with this interpretation, A and D are true.

The case would be much different if the question is “does the function have continuous derivative at $0$?” But “differentiable at $0$” usually means “the function has a derivative at $0$”.
If continuous differentiability is required, then we know that $m>1$ and $f'(0)=0$, in this case. Moreover, for $x\ne0$,
$$
f'(x)=mx^{m-1}\sin\frac{1}{x^n}-nx^{m-n-1}\cos\frac{1}{x^n}
$$
and this is continuous everywhere, except possibly at $0$.
The limit of this function at $0$ should be $0$. Since $m>1$, we already know that
$$
\lim_{x\to0}x^{m-1}\sin\frac{1}{x^n}=0
$$
so we also need that
$$
\lim_{x\to0}x^{m-n-1}\cos\frac{1}{x^n}=0
$$
Note that if $m-n-1\le0$, the limit doesn't exist. Instead, if $m-n-1>0$, the limit is $0$ with the same argument as before.
So the condition for continuous differentiability at $0$ is $m>n+1$.
For example, if $m=2$ and $n=1$, the derivative is
$$
f'(x)=\begin{cases}
2x\sin\dfrac{1}{x}-\cos\dfrac{1}{x} & \text{if $x\ne0$}\\
0 & \text{if $x=0$}
\end{cases}
$$
and this function is not continuous at $0$.
So, if we consider this as the question, we have to choose non differentiability, so either C or D.
A: $f'(x)=mx^{m-1}sin(1/x^n)-nx^{m-n-1}cos(1/x^n)$ (for $x \neq 0$). The sine and cosine function are continuous; therefore differentiability arises if there are factors $x^a$ with $a \geq 0$. Otherwise the derivative of $f(x)$ would have a jump at $x=0$.
