Chapter 6 - Inverse Functions - Review - Exercises - Page 505: 41

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

Given $$y=\ln\left(\frac{1}{x}\right)+\frac{1}{\ln x}$$ Then \begin{align*} y'&=x\left(-\frac{1}{x^2}\right)-\frac{1}{(\ln x)^2} \frac{1}{x}\\ &=-\frac{1}{x} -\frac{1}{(\ln x)^2} \frac{1}{x} \end{align*}