You might want to check out Faa di Bruno's formula: http://mathworld.wolfram.com/FaadiBrunosFormula.html
In your case you have $f(x) = g(h(x))$ where $g(x) = 1/x$ and $h(x) = \cos(x)$. The answer is not simple unfortunately, and involves some combinatorics.
Alternatively, notice $f(x) \cos(x) = 1$, and thus by taking the $m$th derivative of both sides we get
$$ \sum_{k=0}^m {m \choose k} f^{(k)}(x) \left( \frac{d^{m-k}}{dx^{m-k}} \cos(x) \right) = 0 $$
if $m \geq 1$. Thus, you can view the vector $(f(x),f'(x),\dots,f^{(n)}(x))$ as the solution to the system of linear equations above with $m=0,1,\dots,n$. You can invert this matrix numerically, if that if your interest. A closed form might be possible too.