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

$$y'=\frac{1}{t}$$
$$y=\log_2(8t^{\ln2})$$ Recall Theorem 7: $$\frac{d}{dt}\log_au=\frac{1}{u\ln a}\frac{du}{dt}$$ So the derivative of $y$ is: $$y'=\frac{1}{8t^{\ln2}\ln2}(8t^{\ln2})'$$ Remember that $(x^n)'=nx^{n-1}$ $$y'=\frac{(8\ln2)t^{(\ln2)-1}}{8t^{\ln2}\ln2}=\frac{t^{(\ln2)-1}}{t^{\ln2}}$$ $$y'=t^{\ln2-1-\ln2}=t^{-1}$$ $$y'=\frac{1}{t}$$