Answer
$$y'=-\sin\theta\ln(2\theta+2)+\frac{\cos\theta}{\theta+1}$$
Work Step by Step
$$y=(\cos\theta)\ln(2\theta+2)$$
Recall the following Derivative Rules: $$\frac{d}{d\theta}(\ln u)=\frac{1}{u}\frac{du}{d\theta}$$ $$\frac{d}{d\theta}(\cos\theta)=-\sin\theta$$
Therefore, we have $$y'=\Big((\cos\theta)\ln(2\theta+2)\Big)'=(\cos\theta)'\ln(2\theta+2)+\cos\theta\Big(\ln(2\theta+2)\Big)'$$
$$y'=-\sin\theta\ln(2\theta+2)+\cos\theta\times\frac{1}{2\theta+2}(2\theta+2)'$$
$$y'=-\sin\theta\ln(2\theta+2)+\cos\theta\times\frac{2}{2\theta+2}$$
$$y'=-\sin\theta\ln(2\theta+2)+\cos\theta\times\frac{1}{\theta+1}$$
$$y'=-\sin\theta\ln(2\theta+2)+\frac{\cos\theta}{\theta+1}$$
