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

$$y'=\frac{1}{x\ln x\ln(\ln x)}$$
$$y=\ln\Big(\ln(\ln x)\Big)$$ Recall the following Derivative Rules: $$\frac{d}{dx}(\ln u)=\frac{1}{u}\frac{du}{dx}$$ We have $$y'=\Big[\ln\Big(\ln(\ln x)\Big)\Big]'=\frac{1}{\ln(\ln x)}\Big(\ln(\ln x)\Big)'$$ $$y'=\frac{1}{\ln(\ln x)}\times\frac{1}{\ln x}(\ln x)'$$ $$y'=\frac{1}{\ln(\ln x)}\times\frac{1}{\ln x}\times\frac{1}{x}$$ $$y'=\frac{1}{x\ln x\ln(\ln x)}$$