Differentiation of $o(x^n)$ I was investigating the properties of $o(x^n)$ and found that under some circumstances its derivative is $o(x^{n-1})$, but under some other circumstances it is not true, like the function 
$$
x^3\sin \frac{1}{x} (x\neq 0); 0(x=0).
$$
What are the conditions for $o(x^n)’=o(x^{n-1})$ to hold?
Add: Could you provide some other examples of $f(x)$ such that $f(x)=o(x^n)$ for some $n$ but $f’(x)\neq o(x^{n-1})$?
 A: From the definition of $f(x) = o_{x\to a}(g(x))$, you have $$\lim_{x\to a} \frac{f(x)}{g(x)}=0.$$
This means that $$f(x) = \epsilon(x) g(x)$$ with some function $\epsilon$ so that  $\epsilon(x) \to 0$ when $x\to a$.
Back to your problem, let $f(x) = o(x^n)$. By the previous remark, 
$$f(x) = \epsilon(x)x^n$$
with $\lim_{x\to 0} \epsilon(x)= 0$
From my understanding, to achieve the property you want, you need to know more about $\epsilon$ : in the general case, it may not be possible. The first condition you need is that $\epsilon$ is differentiable. Assuming this, 
$$f'(x) = \epsilon'(x)x^n + nx^{n-1}\epsilon(x)$$
To have $f'(x) = o(x^{n-1})$, we must ensure that $\epsilon'(x)x^n = o(x^{n-1})$ itself, i.e. $x\epsilon'(x) \to 0$.
As a conclusion and from my point of view, $o(x^n)' = o(x^{n-1})$ requires that (i) underlying $\epsilon$ is derivable and (ii) $x\epsilon'(x) \to 0$.
These are two strong conditions that might explain why such a situation is rather rare.
A: Recall that by definition 
$$o(x^n)=\omega(x)\cdot x^n$$
with $\omega(x)\to 0$ therefore 
$$\frac{d(o(x^n))}{dx}=\omega’(x)\cdot x^n+ \omega(x)\cdot nx^{n-1}=\left(\omega’(x)\cdot x+ n\omega(x)\right)\cdot x^{n-1}$$
therefore we need that $\exists \omega(x)$ differentiable such that
$$\omega’(x)\cdot x\to 0$$
Your example works indeed since
$$x^3\sin \frac{1}{x}=\frac{\sin \frac1x}{\frac 1x}\cdot x^2 =o(x^2)$$
with $\omega(x)=\frac{\sin \frac1x}{\frac 1x}$ and
$$\omega'(x)\cdot x=\frac{\sin \frac1x}{\frac 1x}-\cos \frac1x$$
