#### Provide solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 14

$\frac{-cosec\: x\frac{4\theta }{2}}{4a}$

Hint:

You must know about derivative of

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

Given:

\begin{aligned} &\text { If } x=a(\theta-\sin \theta), y=a(1+\cos \theta) \\ &\text { Find } \frac{d^{2} y}{d x^{2}} \end{aligned}

Solution:

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

\begin{aligned} &\frac{d x}{d \theta}=a[1-\cos \theta] \; \; \; \; \; ......(1)\\ &\frac{d y}{d \theta}=a(-\sin \theta)\; \; \; \; \; ......(2) \\ &\text { Now } \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{\sin \theta}{(1-\cos \theta)} \end{aligned}

Using identity

\begin{aligned} &\sin 2 \theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \\ &2 \sin ^{2} \frac{\theta}{2}=1-\cos \theta \\ &\cos 2 \theta=1-2 \sin ^{2} \theta \\ &2 \sin ^{2} \theta=1-\cos 2 \theta \end{aligned}

\begin{aligned} &\frac{d y}{d x}=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}} \\ &\frac{d y}{d x}=\cot \frac{\theta}{2} \end{aligned}

Now diff on both sides

\begin{aligned} &\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} \\ &\frac{d^{2} y}{d x^{2}}=-\operatorname{cosec}^{2} \theta \cdot \frac{1}{2} \bullet \frac{1}{a(1-\cos \theta)} \\ &\frac{d^{2} y}{d x^{2}}=\operatorname{cosec}^{2} \frac{\theta}{2} \bullet \frac{1}{2} \bullet \frac{1}{a 2 \sin ^{2} \frac{\theta}{2}} \end{aligned}

\begin{aligned} &=\frac{-\cos e c^{2} \frac{\theta}{2} \times \cos e c^{2} \frac{\theta}{2}}{4 a} \\&\frac{d^{2} y}{d x^{2}}=\frac{\cos e c^{4} \frac{\theta}{2}}{4 a} \end{aligned}

Hence proved