735. Find $r^{\prime}$, if

$$
r=(\sin \varphi)^{\cos 2 \varphi}
$$
Solution. $\ln r=\cos 2 \varphi \ln \sin \varphi ; \frac{r^{\prime}}{r}=(\cos 2 \varphi)^{\prime} \ln \sin \varphi+$ $+\cos 2 \varphi(\ln \sin \varphi)^{\prime}=-2 \sin 2 \varphi \ln \sin \varphi+\cos 2 \varphi \cdot \frac{\cos \varphi}{\sin \varphi}$, hence

$$
r^{\prime}=(\sin \varphi)^{\cos 2 \varphi}(\operatorname{ctg} \varphi \cos 2 \varphi-2 \sin 2 \varphi \ln \sin \varphi) .
$$