Showing that $x^a \sin(1/x), x>0; 0, x=0$ is differentiable if $a>1$ I'm a little confused on how to show something is differentiable. I'm trying to show for the function:
$f_a(x) = \begin{cases}
x^a\sin(1/x)& x>0\\
0& x=0
\end{cases}
$
is differentiable if $x>a$. Do I simply need to show that if $x>a$ then 
$f'_a(x) = \begin{cases}
ax^{a+1}\sin(1/x)& x>0\\
0& x=0
\end{cases}
$
is continuous?
 A: No. You may just consider, as $x \to 0$, $x>0$,
$$
\frac{f_a(x)-f_a(0)}{x-0}=\frac{x^a\sin(1/x)-0}{x-0}=x^{a-1}\sin(1/x).
$$ Does it admit a limit? One may recall that $a>1$.
A: That isn't the derivative for $x>0$. But for $x>0$ you can use standard
rules to prove differentiability and find the derivative.
This is not possible at $x=0$. There you need to go back to the definition of derivative and ask does
$$\lim_{h\to0}\frac{f(0+h)-f(0)}h$$
exist. In this case
$$\frac{f(0+h)-f(0)}h=h^{a-1}\sin(1/h)$$
for $h\ne0$. Near $0$, $\sin(1/h)$ oscillates wildly, but always remains between $-1$ and $1$. So
$$|h^{a-1}\sin(1/h)|\le|h|^{a-1}.$$
For $a>1$, $|h|^{a-1}$ tends to zero as $h\to0$, and this is enough to
show that $f$ is differentiable at $0$.
A: Show that the derivative does not exist as you approach the origin.
$$
 \lim_{x\to 0^+} f'(x) \ne \lim_{x\to0^-} f'(x)
$$
The derivative is
$$
\frac{df}{dx} = 
a x^{a-1} \sin \left(\frac{1}{x}\right)-x^{a-2} \cos \left(\frac{1}{x}\right)
$$

A sequence of plots follow with $a=\frac{1}{2},1,2$.
$$
 f(x) = x^{a} \sin \frac{1}{x}
$$




