Answer
$$y'=\frac{\sin(\log_7\theta)\ln7+\cos(\log_7\theta)}{\ln7}$$
Work Step by Step
$$y=\theta\sin(\log_7\theta)$$
The derivative of $y$ is $$y'=\sin(\log_7\theta)+\theta\cos(\log_7\theta)(\log_7\theta)'$$
Recall Theorem 7: $$\frac{d}{dx}\log_au=\frac{1}{u\ln a}\frac{du}{dx}$$
That means: $$(\log_7\theta)'=\frac{1}{\theta\ln7}(\theta)'=\frac{1}{\theta\ln7}$$
Therefore, $$y'=\sin(\log_7\theta)+\theta\cos(\log_7\theta)\times\frac{1}{\theta\ln7}$$
$$y'=\sin(\log_7\theta)+\frac{\cos(\log_7\theta)}{\ln7}$$
$$y'=\frac{\sin(\log_7\theta)\ln7+\cos(\log_7\theta)}{\ln7}$$
