Question

If $x=a{\sec }^2\theta$, $y=a{\tan }^3\theta$ then $\frac{d^{3}y}{d{x}^3}$

• A. $\frac{-3}{8a^2}{\cot }^3\theta$
• B. `$\frac{3}{8a^2}{\cot }^3\theta$
• C. $3{\sec }^2\theta \tan \theta$
• D. $\frac{3}{4a^2}{\cot }^3\theta$

Answer 1

The correct option is C $3{\sec }^2\theta \tan \theta$

$\alpha =a{\sec }^2\theta ,y=a\tan 3\theta ,\frac{d^{3}y}{d{x}^3}=?$

$\frac{d^{3}y}{d{x}^3}={\frac{}{d}}^3\theta d{\theta }^3\frac{d^3\theta }{d{x}^3}$

$\frac{dy}{d\theta }=\frac{d}{d\theta }\left(a{\tan }^3\theta \right)=a.3{\tan }^2\theta {\sec }^2\theta$

$\frac{d^{2}y}{d{\theta }^2}=a^3({\sec }^2\theta .\tan \theta .{\sec }^2\theta +{\tan }^2\theta .2\sec \theta (\sec \theta \tan \theta \left)\right)$

$=3a(2\tan \theta {\sec }^4\theta +2{\tan }^3\theta {\sec }^2\theta )$

$\frac{dy}{d{\theta }^2}=60\left[\right(\tan \theta .4{\sec }^3\theta \sec \theta +{\sec }^4\theta .{\sec }^2\theta )+({\tan }^3\theta .2\sec \theta \sec \theta \tan \theta +{\sec }^2\theta .3{\tan }^2\theta {\sec }^2\theta \left)\right]$

$=60[5{\sec }^4\theta +{\sec }^6\theta +2{\tan }^4\theta {\sec }^2\theta +3{\tan }^2\theta {\sec }^2\theta ]$

$=60[7{\sec }^4\theta {\tan }^2\theta +2{\tan }^4\theta {\sec }^2\theta +{\sec }^6\theta ]$

$\frac{dx}{d\theta }\Rightarrow \frac{d}{d\theta }\left(a{\sec }^2\theta \right)=a.2\sec \theta \sec \theta \tan \theta$

$=2a\left({\sec }^2\theta \tan \theta \right)$

$\frac{d^{2}x}{d\theta }=2a(\tan \theta .2\sec \theta \tan \theta +{\sec }^2\theta -{\sec }^2\theta )$

$=2a\left(\right(2{\tan }^2\theta {\sec }^2\theta )+{\sec }^4\theta )$

$\frac{d^{3}x}{d{\theta }^3}=2a\left[\right(2{\tan }^2\theta 32\sec \theta \sec \theta \tan \theta +2{\sec }^2\theta +2{\sec }^2\theta 2\tan \theta {\sec }^2\theta )+4{\sec }^3\theta \sec \theta \tan \theta ]$

$=2a[4{\tan }^3\theta {\sec }^2\theta +4\sec \theta {\sec }^4\theta +4\tan \theta {\sec }^3\theta ]$

$=2a.4[{\tan }^3\theta {\sec }^2\theta +2\tan \theta {\sec }^4\theta ]$

$\frac{d^{3}y}{d{x}^3}\Rightarrow \frac{6a(7{\sec }^4\theta {\tan }^2\theta +2{\tan }^4\theta {\sec }^2\theta +{\sec }^6\theta )}{8a\left(\tan \theta {\sec }^2\theta \right[{\tan }^2\theta +2{\sec }^2\theta \left]\right)}$

$\Rightarrow \frac{3}{4}\frac{{\sec }^2\theta [7{\sec }^2\theta {\tan }^2\theta +2{\tan }^4\theta +{\sec }^4\theta ]}{\tan \theta {\sec }^4\theta ({\tan }^2\theta +2{\sec }^2\theta )}$

$\Rightarrow (Since1+{\tan }^2\theta ={\sin }^2\theta )$

$\Rightarrow \frac{3}{4}\left[\frac{7(1+{\tan }^2\theta )\left({\tan }^2\theta \right)+2{\tan }^2\theta +(1+{\tan }^2\theta )}{\tan \theta (\tan \theta +2(1+\tan \theta \left)\right)}\right]$

on further simplyfying we get

$\Rightarrow 3{\sec }^2\theta \tan \theta$