(7) y = 1 + 3 tan 2 θ {\displaystyle y={\sqrt {1+3\tan ^{2}\theta }}} ; y = ( 1 + 3 tan 2 θ ) 1 2 {\displaystyle y=(1+3\tan ^{2}\theta )^{\frac {1}{2}}} .
Let 3 tan 2 θ = v {\displaystyle 3\tan ^{2}\theta =v} .
y = ( 1 + v ) 1 2 {\displaystyle y=(1+v)^{\frac {1}{2}}} ; d y d v = 1 2 1 + v {\displaystyle \quad {\frac {dy}{dv}}={\frac {1}{2{\sqrt {1+v}}}}\;} (see p. 68);
d v d θ = 6 tan θ sec 2 θ {\displaystyle {\frac {dv}{d\theta }}=6\tan \theta \sec ^{2}\theta }
(for, if tan θ = u {\displaystyle \tan \theta =u} ,
v = 3 u 2 ; d v d u = 6 u ; d u d θ = sec 2 θ ; h e n c e d v d θ = 6 ( tan θ sec 2 θ ) h e n c e d y d θ = 6 tan θ sec 2 θ 2 1 + 3 tan 2 θ . {\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}}}
(8) y = sin x cos x {\displaystyle y=\sin x\cos x} .
d y d x = sin x ( − sin x ) + cos x × cos x = cos 2 x − sin 2 x {\displaystyle {\begin{aligned}{\frac {dy}{dx}}&=\sin x(-\sin x)+\cos x\times \cos x\\&=\cos ^{2}x-\sin ^{2}x\end{aligned}}} .
Exercises XIV. (See page 261 for Answers.)
(1) Differentiate the following:
(i) y = A sin ( θ − π 2 ) . (ii) y = sin 2 θ ; and y = sin 2 θ . (iii) y = sin 3 θ ; and y = sin 3 θ . {\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}}}
(2) Find the value of θ {\displaystyle \theta } for which sin θ × cos θ {\displaystyle \sin \theta \times \cos \theta } is a maximum.
(3) Differentiate y = 1 2 π cos 2 π n t {\displaystyle y={\dfrac {1}{2\pi }}\cos 2\pi nt} .