Page:Calculus Made Easy.pdf/193

(7) $y={\sqrt {1+3\tan ^{2}\theta }}$; $y=(1+3\tan ^{2}\theta )^{\frac {1}{2}}$.

Let $3\tan ^{2}\theta =v$.

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

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

(for, if $\tan \theta =u$,

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

(8) $y=\sin x\cos x$.

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

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Exercises XIV. (See page 261 for Answers.)

(1) Differentiate the following:

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

(2) Find the value of $\theta$ for which $\sin \theta \times \cos \theta$ is a maximum.

(3) Differentiate $y={\dfrac {1}{2\pi }}\cos 2\pi nt$.