## Calculus (3rd Edition)

$$y' =\frac{1-\ln x}{x^2}.$$
Recall that $(\ln x)'=\dfrac{1}{x}$ Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Since $y=\frac{\ln x}{x}$, then by the quotient rule, we have $$y'=\frac{x(1/ x)-\ln x}{x^2}=\frac{1-\ln x}{x^2}.$$