#### Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 12 maths

$\frac{32}{27a}$

Hint:

You must know about derivative of

$\frac{cos^{3}\theta}{sin^{3}\theta }$

Given:

\begin{aligned} &{\text { If } x=a\left(1-\cos ^{3} \theta\right), y=a \sin ^{3} \theta} \\ &\text { Prove that } \frac{d^{2} y}{d x^{2}}=\frac{32}{27 a} \; a t\; \theta=\frac{\pi}{6} \end{aligned}

Solution:

\begin{aligned} &{\text { Let } x=a\left(1-\cos ^{3} \theta\right), y=a \sin ^{3} \theta} \end{aligned}

\begin{aligned} &\frac{d y}{d \theta}=3 a \sin ^{2} \theta \cos \theta\left(\frac{d}{d \theta} \sin ^{3} \theta=3 a \sin ^{2} \theta \cos \theta\right) \\ &\frac{d y}{d \theta}=3 \cos ^{2} \theta \sin \theta \ \left(\frac{d}{d \theta} \cos ^{3} \theta=3 \cos ^{2} \theta \sin \theta\right) \\ &\frac{d y}{d x}=\tan \theta \quad \quad\left(\frac{d}{d \theta} \tan \theta=\sec ^{2} \theta\right) \end{aligned}

\begin{aligned} &\frac{\frac{d}{d \theta}\left(\frac{d y}{d x}\right)}{\frac{d x}{d \theta}}=\frac{\sec ^{2} \theta}{3 a \cos ^{2} \theta \sin \theta}\\ &\frac{d^{2} y}{d x^{2}}=\frac{\sec ^{4} \theta}{3 a \sin \theta} \quad\left(\theta=\frac{\pi}{6}\right)\\ &\frac{d^{2} y}{d x^{2}}=\frac{\sec ^{4}\left(\frac{\pi}{6}\right)}{3 a \sin \left(\frac{\pi}{6}\right)}=\frac{32}{27 a} \end{aligned}

Hence proved