Answer
$$y'=\frac{\cos^2\theta-\sin^2\theta}{\ln7\sin\theta\cos\theta}-\frac{1+\ln2}{\ln7}$$
Work Step by Step
$$y=\log_7\Big(\frac{\sin\theta\cos\theta}{e^\theta2^\theta}\Big)$$
Recall Theorem 7: $$\frac{d}{dx}\log_au=\frac{1}{u\ln a}\frac{du}{dx}$$
So the derivative of $y$ is: $$y'=\frac{1}{\frac{\sin\theta\cos\theta}{e^\theta2^\theta}\times\ln7}\Big(\frac{\sin\theta\cos\theta}{e^\theta2^\theta}\Big)'=\frac{e^\theta2^\theta}{\ln7\sin\theta\cos\theta}\Big(\frac{\sin\theta\cos\theta}{e^\theta2^\theta}\Big)'$$
We examine $(\frac{\sin\theta\cos\theta}{e^\theta2^\theta})'$: $$\Big(\frac{\sin\theta\cos\theta}{e^\theta2^\theta}\Big)'=\frac{(\sin\theta\cos\theta)'(e^\theta2^\theta)-(\sin\theta\cos\theta)(e^\theta2^\theta)'}{(e^\theta2^\theta)^2}$$ $$=\frac{(\cos\theta\cos\theta+\sin\theta(-\sin\theta))(e^\theta2^\theta)-(\sin\theta\cos\theta)(e^\theta2^\theta+e^\theta2^\theta\ln2)}{(e^\theta2^\theta)^2}$$ (since $(a^x)'=a^x\ln a$) $$=\frac{(\cos^2\theta-\sin^2\theta)(e^\theta2^\theta)-(\sin\theta\cos\theta)(1+\ln2)(e^\theta2^\theta)}{(e^\theta2^\theta)^2}$$ $$=\frac{(\cos^2\theta-\sin^2\theta)-(\sin\theta\cos\theta)(1+\ln2)}{e^\theta2^\theta}$$
Therefore, $$y'=\frac{e^\theta2^\theta}{\ln7\sin\theta\cos\theta}\times\frac{(\cos^2\theta-\sin^2\theta)-(\sin\theta\cos\theta)(1+\ln2)}{e^\theta2^\theta}$$ $$y'=\frac{(\cos^2\theta-\sin^2\theta)-(\sin\theta\cos\theta)(1+\ln2)}{\ln7\sin\theta\cos\theta}$$ $$y'=\frac{\cos^2\theta-\sin^2\theta}{\ln7\sin\theta\cos\theta}-\frac{1+\ln2}{\ln7}$$
