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.  
 A: The derivative of $x^n\log x$ is $x^{n-1}(n\log x+1)$ by the product rule.
The first derivative is: $x^7+8x^7\log x$.
Edit: For further steps note that we can ignore the $x^n$ terms (they will die out after $9$ derivatives) So the second derivative (with the $7x^6$ ignored) is:
$8 \cdot x^6(7\log x +1)$
Ignoring the $x^n$ term here again yields $56x^6 \log x$
Since this will go on $9$ times, we will eventually have to take the derivative of $8!\log x$ which will yield $\frac{8!}{x}$
As noted in the comments, this method would be cumbersome if the $x^n$ terms did not die out.
A: Notice
$$\require{cancel}
\frac{\partial^9}{\partial x^9}\left[ x^\beta\log x\right]
= \frac{\partial^{10}}{\partial x^9\partial\beta}\left[x^\beta\right]
= \frac{\partial^{10}}{\partial\beta\partial x^9}\left[x^\beta\right]
= \frac{\partial}{\partial\beta}\left[\prod_{k=0}^8(\beta-k)x^{\beta-9}\right]
= \frac{\partial}{\partial\beta}\left[\prod_{k=0}^8(\beta-k)\right]x^{\beta-9}
+\left[\prod_{k=0}^8(\beta-8)\right] x^{\beta-9}\log x$$
Taking $\beta = 8$ and throw away terms that will vanish at $\beta = 8$, above expression
becomes
$$
\left\{
\color{red}{\cancelto{0}{\color{silver}{\frac{\partial}{\partial\beta}\left[\prod_{k=0}^7(\beta-k)\right](\beta-8)}}}
+ \left[\prod_{k=0}^7(\beta-k)\right]\right\}x^{\beta-9}
+
\color{red}{\cancelto{0}{\color{silver}{\left[\prod_{k=0}^8(\beta-8)\right] x^{\beta-9}}}}\log x
$$
and hence
$$\frac{d^9}{dx^9}\left[x^8\log x \right] = \frac{8!}{x}\;$$
A: 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 
$$D^9(x^8\log(x))=\sum_{n=0}^\infty \binom{9}{n}(D^{9-n}x^8)D^nlog(x)$$
$$=\sum_{n=1}^\infty \binom{9}{n}8!\frac{x^{n-1}}{(n-1)!}(-1)^{n+1}\frac{(n-1)!}{x^n}=\frac{8!}{x}\sum_{n=1}^\infty \binom{9}{n}(-1)^{n+1}=\frac{8!}{x}(1-(1-1)^9)=\frac{8!}{x}.$$
A: Let us define $$D_n=\frac{d^n}{dx^n}(x^8\ln x)$$ and follow what illysial suggested. You then have $$D_1=x^7+8 x^7 \log (x)$$ $$D_2=15 x^6+56 x^6 \log (x)$$ $$D_3=146 x^5+336 x^5 \log (x)$$ $$D_4=1066 x^4+1680 x^4 \log (x)$$ and so on. When going to $D_9$, the derivative of the polynomial part is zero and the only thing which remains in $D_8$ is $$8 \times 7\times 6\times 5\times 4\times 3\times 2 \times 1 \times \log(x)$$ and then illysial's result. 
A: 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. 
