Calculus (3rd Edition)

$$f'(\theta)= \frac{ 1}{\theta} \cos(\ln\theta).$$
Recall that $(\ln x)'=\dfrac{1}{x}$ Recall that $(\sin x)'=\cos x$. Since $f(\theta)=\sin ( \ln\theta)$, then the derivative, using the chain rule, is given by $$f'(\theta)=\cos( \ln \theta) (\ln \theta)'= \cos (\ln\theta )\frac{ 1}{\theta}=\frac{ 1}{\theta} \cos(\ln\theta).$$