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  4. If x = 1 sec2 θ , y = a tan2 θ then (d2 y/dx2) =

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Q. If x = 1 $\sec^2 \; \theta ,$ y = a $\tan^2 \theta$ then $\frac{d^2 y}{dx^2}$ =

KCETKCET 2019Continuity and Differentiability

Solution:

$\frac{d x}{d \theta}=2a \, sec^{2} \,\theta \, tan\,\theta$
$\frac{d y}{d\theta}=2a \, tan\, \theta \, sec^{2}\theta$
$\therefore \, \frac{d\,y}{d\,x}=\frac{d \,y}{\frac{d \,\theta}{\frac{d \,x}{d \,\theta}}}=\frac{2a\, tan\,\theta. sec^{2}\,\theta}{2a \, tan\,\theta \, sec^{2}\,\theta}=1\quad\therefore \, \frac{d^{2}y}{dx^{2}}=0$