TITLE: How to prove this inequality for $n$-th derivative: $\left|\frac{d^n}{dx^n}\left(\frac{\sin x}{x}\right)\right|\leq \frac1n$?
QUESTION [7 upvotes]: It's easy to see that $\frac{d^n(\sin x)}{dx^n}=\sin \left(x+\frac{\pi n}{2}\right)$, so the following inequality holds: $$\left|\frac{d^n(\sin x)}{dx^n}\right|=\left|\sin \left(x+\frac{\pi n}{2}\right)\right|\leq 1.$$
I'm interested in proving a similar inequality for the function $\frac{\sin x}{x}$. I suspect that for all $n$
$$\left|\frac{d^n}{dx^n}\left(\frac{\sin x}{x}\right)\right|\leq \frac1n.$$
Unfortunately I don't know any nice expression for this $n$-th derivative which could help. I'll be grateful for all useful suggestions.

REPLY [14 votes]: A hint:
$${\sin x\over x}=\int_0^1\cos(t\,x)\ dt\ .$$