Chapter 2 - The Derivative - 2.3 Introduction To Techniques Of Differentiation - Exercises Set 2.3 - Page 142: 78

$f'(x) = -\frac{1}{(x-1)^2}$

The result of Exercise $75$ says that given $f(x) = (mx+b)^{n}$, then $f'(x) = nm(mx+b)^{n-1}$ Applying this result to $f(x) = \frac{1}{x-1} = (x-1)^{-1}$ where $m=1$, $n=-1$, and $b=-1$, we get: $f'(x) = nm(mx+b)^{n-1} = (-1)(1)(x-1)^{((-1)-1)} = -(x-1)^{-2}= -\frac{1}{(x-1)^2}$