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

$$y'=\frac{3}{(\ln2\ln8)t\log_2t}$$
$$y=3\log_8(\log_2t)$$ Recall Theorem 7: $$\frac{d}{dt}\log_au=\frac{1}{u\ln a}\frac{du}{dt}$$ So the derivative of $y$ is: $$y'=3\Big(\frac{1}{\log_2t\times\ln8}(\log_2t)'\Big)=\frac{3(\log_2t)'}{(\ln8)\log_2t}$$ For $(\log_2t)'$, we have $$(\log_2t)'=\frac{1}{t\ln2}(t)'=\frac{1}{t\ln2}$$ Therefore, $$y'=\frac{\frac{3}{t\ln2}}{(\ln8)\log_2t}=\frac{3}{(\ln2\ln8)t\log_2t}$$