Calculus (3rd Edition)

$$\frac{d}{d t} \log _{3}(\sin t) =\frac{1}{\ln 3 } \cot t.$$
Recall that $(\log_b x)'=\dfrac{1}{(\ln b)x}$ Recall that $(\sin x)'=\cos x$. Thus we have: $$\frac{d}{d t} \log _{3}(\sin t)=\frac{\cos t}{ (\ln 3) \sin t }=\frac{1}{\ln 3 } \cot t.$$ Since $\cot t=\dfrac{\cos t}{\sin t}$.