## Calculus: Early Transcendentals 8th Edition

$\frac{d^{9}}{dx^{9}}(x^{8}lnx)=\frac{8!}{x}$
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, $\frac{d^{9}}{dx^{9}}(x^{8}lnx)=\frac{8!}{x}$