Answer
$$y'=\frac{3^{\log_2t}\ln3}{t\ln2}$$
Work Step by Step
$$y=3^{\log_2t}$$
We have the following theorems $$\frac{d}{dt}\log_au=\frac{1}{u\ln a}\frac{du}{dt}$$ $$\frac{d}{dt}(a^u)=a^u\ln a\frac{du}{dt}$$
So the derivative of $y$ is: $$y'=(3^{\log_2t})'=3^{\log_2t}\ln3(\log_2t)'$$
$$y'=3^{\log_2t}\ln3\times\frac{1}{t\ln2}(t)'$$
$$y'=\frac{3^{\log_2t}\ln3}{t\ln2}$$
