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

$$y'=\frac{\Big(1-\ln(\ln x)\Big)(\ln x)^{1/(\ln x)}}{x\ln^2x}$$
$$y=(\ln x)^{1/(\ln x)}$$ Take the natural logarithm of both sides: $$\ln y=\ln\Big((\ln x)^{1/(\ln x)}\Big)$$ $$\ln y=\frac{1}{\ln x}\times\ln(\ln x)=\frac{\ln(\ln x)}{\ln x}$$ Now take the derivative of both sides, and remember that $(\ln x)'=1/x$: $$\frac{1}{y}\times y'=\frac{\Big(\ln(\ln x)\Big)'\ln x-\Big(\ln(\ln x)\Big)(\ln x)'}{\ln^2x}$$ $$\frac{1}{y}\times y'=\frac{\Big(\frac{(\ln x)'}{\ln x}\Big)\ln x-\frac{\ln(\ln x)}{x}}{\ln^2x}$$ $$\frac{y'}{y}=\frac{\Big(\frac{1}{x\ln x}\Big)\ln x-\frac{\ln(\ln x)}{x}}{\ln^2x}$$ $$\frac{y'}{y}=\frac{\frac{1}{x}-\frac{\ln(\ln x)}{x}}{\ln^2x}$$ $$\frac{y'}{y}=\frac{\frac{1-\ln(\ln x)}{x}}{\ln^2x}=\frac{1-\ln(\ln x)}{x\ln^2x}$$ Find $y'$: $$y'=\Big(\frac{1-\ln(\ln x)}{x\ln^2x}\Big)y$$ $$y'=\frac{\Big(1-\ln(\ln x)\Big)(\ln x)^{1/(\ln x)}}{x\ln^2x}$$