TITLE: Is there an easier way to find $\frac{\mathrm d^9}{\mathrm dx^9}(x^8\ln x)$ than using the product rule repeatedly?
QUESTION [8 upvotes]: Find $\dfrac{\mathrm d^9}{\mathrm dx^9}(x^8\ln x)$.

I know how to solve this problem by repeatedly using the product rule, but I was wondering if there is a short cut. Thanks.

REPLY [2 votes]: Actually, the succinct Leibniz product rule 
$$D^m(f(x)g(x))=\sum_{n=0}^\infty \binom{m}{n}(D^{m-n}f(x))(D^ng(x))$$
works quickly and transparently, so 
$$D^9(x^8\log(x))=\sum_{n=0}^\infty \binom{9}{n}(D^{9-n}x^8)D^nlog(x)$$
$$=\sum_{n=1}^\infty \binom{9}{n}8!\frac{x^{n-1}}{(n-1)!}(-1)^{n+1}\frac{(n-1)!}{x^n}=\frac{8!}{x}\sum_{n=1}^\infty \binom{9}{n}(-1)^{n+1}=\frac{8!}{x}(1-(1-1)^9)=\frac{8!}{x}.$$