Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 25

$$y'=\frac{1-\ln t}{t^2}$$

$$y=\frac{\ln t}{t}$$ Recall the following Derivative Rules: $$\frac{d}{dt}(\ln t)=\frac{1}{t}$$ $$\frac{d}{dt}\Big(\frac{u}{v}\Big)=\frac{v\frac{du}{dt}-u\frac{dv}{dt}}{v^2}$$ Therefore, we have $$y'=\frac{t(\ln t)'-\ln t(t)'}{t^2}=\frac{t\times\frac{1}{t}-\ln t}{t^2}$$ $$y'=\frac{1-\ln t}{t^2}$$