Answer
$$y'=\frac{\log_3r}{r\ln3}$$
Work Step by Step
$$y=\log_{3}r\log_9r=\log_3r\log_{3^2}r$$
We have $\log_a^bx=\frac{1}{b}\log_ax$. Therefore, $\log_{3^2}r=\frac{1}{2}\log_3r$
$$y=\log_3r\times\frac{1}{2}\log_3r=\frac{1}{2}(\log_3r)^2$$
The derivative of $y$ is $$y'=\Big(\frac{1}{2}(\log_3r)^2\Big)'=\frac{1}{2}\times2\log_3r(\log_3r)'=\log_3r(\log_3r)'$$
Recall Theorem 7: $$\frac{d}{dr}\log_au=\frac{1}{u\ln a}\frac{du}{dr}$$
That means: $$y'=\log_3r\times\frac{1}{r\ln3}(r)'=\frac{\log_3r}{r\ln3}$$
