Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.1 Exercises - Page 325: 42

$\displaystyle \frac{1}{x-1}$

By theorem 5.3/2: $\displaystyle \frac{d}{dx}[\ln u]=\frac{1}{u}\cdot\frac{du}{dx}=\frac{u^{\prime}}{u},\ \ \ (u>0)$ Here, u(x)$=x-1,\displaystyle \ \ \ \frac{du}{dx}=1$ $\displaystyle \frac{d}{dx}[\ln(x-1)]=\frac{1}{x-1}\cdot 1=\frac{1}{x-1}$