<s>[INST] Is there an easier way to find $\frac{\mathrm d^9}{\mathrm dx^9}(x^8\ln x)$ than using the product rule repeatedly? 
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.  
 [/INST] For any analytic function $f$, 
$$\sum_{n=0}^\infty  \dfrac{z^n}{n!} \dfrac{d^n}{dx^n} f(x) = f(x+z) $$
Thus $\dfrac{d^n f}{dx^n}$ is $n!$ times the coefficient of $z^n$ in the Maclaurin series of $f(x+z)$.
In this case $f(x+z) = (x+z)^8 \ln(x+z)$.  Now $$
\eqalign{(x+z)^8 &= \sum_{j=0}^8 {8 \choose j} x^{8-j} z^j \cr
\ln(x+z) &= \ln(x) + \ln(1 + z/x) = \ln(x) + \sum_{k=1}^\infty (-1)^{k-1} \dfrac{z^k}{k x^k}\cr}$$
Thus what you want  is
$$\eqalign{ 9! &\left( {8 \choose 0} x^8 \dfrac{1}{9 x^9} - {8 \choose 1} x^7 \dfrac{1}{8 x^8} + \ldots + {8 \choose 8} x^0 \dfrac{1}{1 x^1}\right)\cr
&=\dfrac{9!}{x} \left(\dfrac{1}{9}- \dfrac{8}{8} +  \dfrac{28}{7} - \dfrac{56}{6} + \dfrac{70}{5} - \dfrac{56}{4} + \dfrac{28}{3} - \dfrac{8}{2} + \dfrac{1}{1}\right)  = \dfrac{9!}{9x} = \dfrac{8!}{x} = \dfrac{40320}{x}}$$
How did this work out so nicely?  You might notice that 
$$ { 8 \choose j} \dfrac{1}{9-j} = \dfrac{8!}{j! (9-j)!} = \dfrac{1}{9} {9 \choose j}$$
and  the $\dfrac{(-1)^j}{9} {9 \choose j}$ and $\dfrac{(-1)^{9-j}}{9} {9 \choose 9-j}$ terms cancel out for $j = 1 \ldots 4$, leaving just the $j=0$ term. 
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