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

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