Answer
$$y'=2\cos(\ln\theta)$$
Work Step by Step
$$y=\theta(\sin(\ln\theta)+\cos(\ln\theta))$$
Recall the following Derivative Rules: $$\frac{d}{d\theta}(\ln\theta)=\frac{1}{\theta}$$ $$\frac{d}{d\theta}(uv)=v\frac{du}{d\theta}+u\frac{dv}{d\theta}$$ $$\frac{d}{d\theta}(u+v)=\frac{du}{d\theta}+\frac{dv}{d\theta}$$
We have $$y'=\Big(\theta(\sin(\ln\theta)+\cos(\ln\theta))\Big)'$$
$$y'=(\theta)'(\sin(\ln\theta)+\cos(\ln\theta))+\theta(\sin(\ln\theta)+\cos(\ln\theta))'$$
$$y'=(\sin(\ln\theta)+\cos(\ln\theta))+\theta\Big[(\sin(\ln\theta))'+(\cos(\ln\theta))'\Big]$$
$$y'=(\sin(\ln\theta)+\cos(\ln\theta))+\theta\Big[\cos(\ln\theta)(\ln\theta)'-\sin(\ln\theta)(\ln\theta)'\Big]$$
$$y'=(\sin(\ln\theta)+\cos(\ln\theta))+\theta\Big[\frac{\cos(\ln\theta)}{\theta}-\frac{\sin(\ln\theta)}{\theta}\Big]$$
$$y'=(\sin(\ln\theta)+\cos(\ln\theta))+[\cos(\ln\theta)-\sin(\ln\theta)]$$
$$y'=2\cos(\ln\theta)$$
