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#### Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 9 maths textbook solution

$\frac{sec^{3}\theta }{a\theta }$

Hint:

You must know about derivative of

$cos\: \theta \: \: and\: \: sin\: \theta$

Given:

\begin{aligned} &\text { If } x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta) \text { Prove that }\\ &\frac{d^{2} x}{d \theta^{2}}=a(\cos \theta-\theta \sin \theta), \frac{d^{2} y}{d \theta^{2}}=a(\sin \theta-\theta \cos \theta), \frac{d^{2} y}{d x^{2}}=\frac{\sec ^{3} \theta}{a \theta} \end{aligned}

Solution:

\begin{aligned} &\text { Let } x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta) \end{aligned}

Use multiplicative rule

\begin{aligned} &\text { As } U V=U V^{1}+U^{1} V \\ &\text { Where } U=\theta \& V=\sin \theta \\ &\begin{array}{l} \frac{d x}{d \theta}=a[-\sin \theta+\theta \cos \theta+\sin \theta] \quad\left[\frac{d \sin \theta}{d x}=\cos \theta, \frac{d \cos \theta}{d x}=-\sin \theta\right] \\ \\ \frac{d x}{d \theta}=a \theta \cos \theta \end{array} \end{aligned}

again

Use multiplicative rule

\begin{aligned} &\text { As } U V=U V^{1}+U^{1} V \\ &\text { Where } U=\theta\: \&\: V=\cos \theta \\ &\frac{d y}{d \theta}=a[\cos \theta+\theta \sin \theta-\cos \theta] \\ &\frac{d y}{d \theta}=a \theta \sin \theta \end{aligned}

Again Use multiplicative rule

\begin{aligned} &\text { As } U V=U V^{1}+U^{1} V \\ &\text { Where } U=\theta \& V=\cos \theta \\ &\frac{d^{2} x}{d \theta^{2}}=a[-\theta \sin \theta+\cos \theta] \\ &\frac{d^{2} x}{d \theta^{2}}=a[\cos \theta-\theta \sin \theta] \end{aligned}

Again Use multiplicative rule

\begin{aligned} &\text { As } U V=U V^{1}+U^{1} V \\ &\text { Where } U=\theta \& V=\sin \theta \\ &\frac{d^{2} y}{d \theta^{2}}=a\{\theta \cos \theta+\sin \theta\} \\ &\frac{d^{2} x}{d \theta^{2}}=a(\cos \theta-\theta \sin \theta) \end{aligned}

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

\begin{aligned} &\frac{d}{d \theta}(\tan \theta) \times \frac{1}{a \theta \cos \theta} \\ &\frac{\sec ^{2} \theta}{a \theta \cos \theta} \\ &\frac{d^{2} y}{d x^{2}}=\frac{\sec ^{3} \theta}{a \theta} \end{aligned}