Answer
$$\frac{dy }{d \theta} =\frac{1}{\ln7}\left[ 2\cot 2 \theta - 1-\ln 2\right]\\ $$
Work Step by Step
Given $$ y =\log _{7}\left(\frac{\sin \theta \cos \theta}{e^{\theta} 2^{\theta}}\right)\\$$
Since $$\log_{a}z=\frac{\ln z}{\ln a}$$ So, we have
\begin{aligned}
y&
=\frac{1}{\ln7}\ln \left(\frac{2 \sin \theta \cos \theta}{2 e^{\theta} 2^{\theta}}\right)\\
& =\frac{1}{\ln7}\ln\left(\frac{\sin 2 \theta}{e^{\theta} 2^{\theta+1}}\right)\\
& =\frac{1}{\ln7}\left[\ln(\sin 2 \theta)-\ln\left(e^{\theta} 2^{\theta+1}\right)\right]\\
& =\frac{1}{\ln7}\left[\ln(\sin 2 \theta)-\ln \left(e^{\theta}\right)-\ln \left(2^{\theta+1}\right)\right]\\
&=\frac{1}{\ln7}\left[\ln(\sin 2 \theta)-\theta \ln e-(\theta+1) \ln 2\right]\\
&\Rightarrow \frac{dy }{d \theta}=\frac{1}{\ln7}\left[\frac{2\cos 2 \theta}{\sin 2 \theta}- \ln e-\ln 2\right]\\
&\ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\ln7}\left[ 2\cot 2 \theta - 1-\ln 2\right]\\
\end{aligned}
