Answer
$f^{n}(x)= \frac{(-1)^{n-1}(n-1)!}{(x-1)^{n}}$
Work Step by Step
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,$f^{n}(x)= \frac{(-1)^{n-1}(n-1)!}{(x-1)^{n}}$
