nth derivative cardinal sine function. I need to find formula for $n$-th derivative of $\mathrm{sinc}(x)$. The following question isn't related to homework but it's a question that seems very challenging to me, and I take some interest in it.
I have  been trying to find the $n$th derivative but cant.
Help is appreciated, thanks.
 A: Write $f(x)=\operatorname{sinc}(x)=\frac{\sin(x)}{x}$.
Let $g(x)=xf(x)=\sin x$. Then $g^{(n)}(x) = xf^{(n)}(x) + \sum_{k=0}^{n-1} f^{(k)}(x)$. We know $g^{(n)}(x)$, so we get:
$$f^{(n)}(x) = \frac{1}{x}\left(g^{(n)}(x)- \sum_{k=0}^{n-1}f^{(k)}(x)\right)$$
This gives a recursive formula for $f^{(n)}(x)$. Not sure if that is helpful.
Letting $h_n(x)=x^n f^{(n)}(x)$, you get that:
$$h_n(x)=x^ng^{(n)}(x) - \sum_{k=0}^{n-1}x^{n-1-k}h_k(x)$$
The nice thing about this is that $h_n(x)$ is of the form $p_n(x)\cos x + q_n(x)\sin x$ where $p_n$ and $q_n$ are polynomials, which might make things easier to combine.
A: Recall the general Liebniz rule: the $n$th derivative of a product $A(x)B(x)$ is given by$$(AB)^{(n)}(x) = \sum_{k=0}^n \binom{n}{k} A^{(k)}(x)B^{(n-k)}(x).$$ Therefore
$$f^{(n)}(x)=\frac{d^n}{dx^n}\frac{\sin x}{x}=\sum_{k=0}^n \binom{n}{k} \frac{d^{n-k}}{dx^{n-k}}\!\!\left(\frac{1}{x}\right)\frac{d^{k}}{dx^{k}}\!\!\left(\sin x\right)$$
The $k$th derivative of $1/x$ yields $(-1)^{k}/k!\,x^{1+k}$, whereas even (odd) derivatives of $\sin x$ give back $\sin x$ up to a sign. So we can write the $n$th derivative as $f^{(n)}(x) = A_n(x) \sin x + B_n(x)\cos x$, and some tedious algebra gives $A_n(x)$ and $B_n(x)$ given as
$$ A_n(x) = \sum_{m=0} \binom{n}{2m} \frac{(-1)^{n+m}}{(2m)!}\frac{1}{x^{2m+1}},\hspace{.5cm} 
   B_n(x) = \sum_{m=0} \binom{n}{2m+1} \frac{(-1)^{n+m-1}}{(2m+1)!}\frac{1}{x^{2m+2}}.$$ Presumably, these are the same as the polynomials in @ThomasAndrews' post apart from an overall factor of $x^{n+1}$.
A: Related problems. You can get a formula as an integral representation. First we write $\frac{\sin(x)}{x}$ as an integral representation  

$$\frac{\sin(x)}{x} = - \int_{0}^{1} \cos(tx) dt. $$

Differentiating both sides of the above equation w.r.t. $x$ we have

$$ \left( \frac{\sin(x)}{x} \right)^{(n)} = - \int_{0}^{1} t^n \cos(tx+n\pi/2) dt. $$

Note: We used the identity 

$$(\cos(at))^{(n)}= a^n \cos(at+n\pi/2) $$

which is not hard to prove.
