Answer
$$
\frac{d}{d t} \log _{3}(\sin t) =\frac{1}{\ln 3 } \cot t.
$$
Work Step by Step
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 $\cot t=\dfrac{\cos t}{\sin t}$.
