## University Calculus: Early Transcendentals (3rd Edition)

$$y'=-\frac{1}{x}$$
$$y=\ln\frac{3}{x}$$ Recall the Derivative Rule for Natural Logarithm: $$\frac{d}{dx}(\ln u)=\frac{1}{u}\frac{du}{dx}$$ Therefore, we have $$y'=\Big(\ln\frac{3}{x}\Big)'=\frac{1}{\frac{3}{x}}\Big(\frac{3}{x}\Big)'=\frac{x}{3}\times\frac{(3)'x-3(x)'}{x^2}$$ $$y'=\frac{1}{3}\times\frac{0\times x-3\times1}{x}=\frac{1}{3}\times\frac{(-3)}{x}$$ $$y'=-\frac{1}{x}$$