Answer
$$y'=\frac{\log_2r}{r\ln2}$$
Work Step by Step
$$y=\log_{2}r\log_4r=\log_2r\log_{2^2}r$$
We have $\log_a^bx=\frac{1}{b}\log_ax$. Therefore, $\log_{2^2}r=\frac{1}{2}\log_2r$
$$y=\log_2r\times\frac{1}{2}\log_2r=\frac{1}{2}(\log_2r)^2$$
The derivative of $y$ is $$y'=\Big(\frac{1}{2}(\log_2r)^2\Big)'=\frac{1}{2}\times2\log_2r(\log_2r)=\log_2r(\log_2r)'$$
Recall Theorem 7: $$\frac{d}{dr}\log_au=\frac{1}{u\ln a}\frac{du}{dr}$$
That means: $$y'=\log_2r\times\frac{1}{r\ln2}(r)'=\frac{\log_2r}{r\ln2}$$
