Answer
$\frac{d^{9}}{dx^{9}}(x^{8}lnx)=\frac{8!}{x}$
Work Step by Step
Find $\frac{d^{9}}{dx^{9}}(x^{8}lnx)$.
Consider $y=(x^{8}lnx)$
$y'=x^{8}\times\frac{1}{8}+8x^{7}lnx$
$y''=15x^{6}+8.7x^{6}lnx$
$y'''=14x^{5}+8.7.6x^{5}lnx$
Proceeding in the same manner, we will have the 9th derivative:
$y^{9'}=\frac{8!}{x}$
Hence, $\frac{d^{9}}{dx^{9}}(x^{8}lnx)=\frac{8!}{x}$
