Answer
$$y'=\frac{1}{t}$$
Work Step by Step
$$y=\log_2(8t^{\ln2})$$
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 $(x^n)'=nx^{n-1}$
$$y'=\frac{(8\ln2)t^{(\ln2)-1}}{8t^{\ln2}\ln2}=\frac{t^{(\ln2)-1}}{t^{\ln2}}$$
$$y'=t^{\ln2-1-\ln2}=t^{-1}$$
$$y'=\frac{1}{t}$$
