#### Provide solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 46

$\frac{8\sqrt{3}}{a}$

Hint:

You must know the derivative of cos, sin, tan and logarithm function

Given:

\begin{aligned} &x=a\left(\cos t+\log \tan \frac{t}{2}\right) \\ &y=a \sin t \quad \text { find } \frac{d^{2} y}{d x^{2}} \text { at } t=\frac{\pi}{3} \end{aligned}

Solution:

\begin{aligned} &x=a\left(\cos t+\log \tan \frac{t}{2}\right) \\ &\frac{d x}{d t}=a\left[-\sin t+\frac{1}{\tan \frac{t}{2}} \sec ^{2} \frac{t}{2} \cdot \frac{1}{2}\right] \\ &\frac{d x}{d t}=a\left[-\sin t+\frac{\cos \frac{t}{2}}{2 \sin \frac{t}{2}} \frac{1}{\cos ^{2} \frac{t}{2}}\right] \end{aligned}

\begin{aligned} &\frac{d x}{d t}=a\left[-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}\right] \\ &\frac{d x}{d t}=a\left[-\sin t+\frac{1}{\sin t}\right] \\ &\frac{d x}{d t}=\frac{a \cos ^{2} t}{\sin t} \\ &y=a \sin t \end{aligned}

\begin{aligned} &\frac{d y}{d t}=a \cos t \\ &\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{a \cos t}{a \cos ^{2} t} \times \sin t=\tan t \\ &\text { Again } \frac{d^{2} y}{d x^{2}}=\sec ^{2} t \times \frac{\sin t}{a \cos ^{2} t} \end{aligned}

\begin{aligned} &\left.\frac{d^{2} y}{d x^{2}}\right]_{t=\frac{\pi}{3}}=\frac{\sec ^{2} \frac{\pi}{3} \cdot \sin \frac{\pi}{3}}{a \cos ^{2} \frac{\pi}{4}}=\frac{(2)^{2} \cdot\left(\frac{\sqrt{3}}{2}\right)}{a\left(\frac{1}{2}\right)^{2}} \\ &\left.\frac{d^{2} y}{d x^{2}}\right]_{t=\frac{\pi}{3}}=\frac{8 \sqrt{3}}{a} \end{aligned}