${\displaystyle y=(1+v)^{\frac {1}{2}}}$; ${\displaystyle \quad {\frac {dy}{dv}}={\frac {1}{2{\sqrt {1+v}}}}\;}$ (see p. 68);

${\displaystyle {\frac {dv}{d\theta }}=6\tan \theta \sec ^{2}\theta }$

${\displaystyle {\begin{aligned}v&=3u^{2};\quad {\frac {dv}{du}}=6u;\quad {\frac {du}{d\theta }}=\sec ^{2}\theta$ ;\\hence\quad \quad \quad {\frac {dv}{d\theta }}&=6(\tan \theta \sec ^{2}\theta )\\hence\quad \quad \quad {\frac {dy}{d\theta }}&={\frac {6\tan \theta \sec ^{2}\theta }{2{\sqrt {1+3\tan ^{2}\theta }}}}.\end{aligned}}}

${\displaystyle {\begin{aligned}{\frac {dy}{dx}}&=\sin x(-\sin x)+\cos x\times \cos x\\&=\cos ^{2}x-\sin ^{2}x\end{aligned}}}$.

${\displaystyle {\begin{aligned}{\text{(i)}}\quad y&=A\sin \left(\theta -{\frac {\pi }{2}}\right).\\{\text{(ii)}}\quad y&=\sin ^{2}\theta$ ;\quad {\text{and }}y=\sin 2\theta .\\{\text{(iii)}}\quad y&=\sin ^{3}\theta ;\quad {\text{and }}y=\sin 3\theta .\end{aligned}}}