## Calculus: Early Transcendentals 8th Edition

$f^{n}(x)= \frac{(-1)^{n-1}(n-1)!}{(x-1)^{n}}$
Differentiate $f(x)=ln(x-1)$ with respect to $x$. $f'(x)=\frac{1}{(x-1)}$ $f''(x)=(-1)(x-1)^{-2}$ $f'''(x)=(-1)^{2}1.2(x-1)^{-3}$ After solving in the same manner, we find that $f^{n}(x)=(-1)^{n-1}1.2.3....(n-1)(x-1)^{-n}$ Hence,$f^{n}(x)= \frac{(-1)^{n-1}(n-1)!}{(x-1)^{n}}$