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### Answers (1)

Answer:

$2 \sqrt{2}=\frac{d^{2} y}{d x^{2}} \text { and } \frac{-1}{\sqrt{x}}=\frac{d^{2} y}{d x^{2}}$

Hint:

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

Given:

\begin{aligned} &x=\cos t+\log \tan \frac{t}{2} \\ &y=\sin t \quad \text { find } \frac{d^{2} y}{d x^{2}} \& \frac{d^{2} y}{d x^{2}} \quad \text { at } t=\frac{\pi}{4} \end{aligned}

Solution:

\begin{aligned} &y=\sin t \quad \frac{d y}{d t}=\cos t \\ &\frac{d^{2} y}{d t^{2}}=-\sin t \\ &\left.\frac{d^{2} y}{d t^{2}}\right]_{t=\frac{\pi}{4}}=-\sin \frac{\pi}{4}=\frac{-1}{\sqrt{2}} \end{aligned}

\begin{aligned} &\text { Again differentiating }\\ &x=\cos t+\log \tan \frac{t}{2}\\ &\frac{d x}{d t}=-\sin t+\frac{1}{\tan \frac{t}{2}} \cdot \sec ^{2} \frac{t}{2} \cdot \frac{1}{2}\\ &=-\sin t+\frac{\cos \frac{t}{2}}{2 \times \sin \frac{t}{2}} \cdot \sec ^{2} \frac{t}{2} \cdot \frac{1}{2}\\ &=-\sin t+\frac{1}{\sin 2 \times \frac{t}{2}} \end{aligned}

\begin{aligned} &=-\sin t+\operatorname{cosec\: t} \\ &\text { Now }, \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{\cos t}{\operatorname{cosec\: t}-\sin t} \\ &=\frac{\cos t}{1-\sin ^{2} t} \sin t=\frac{\sin t \cos t}{\cos ^{2} t} \\ &\frac{d^{2} y}{d x^{2}}=\frac{d\left(\frac{d y}{d x}\right)}{d x} \end{aligned}

\begin{aligned} &=\frac{\frac{d}{d t}\left(\frac{d y}{d x}\right)}{\frac{d x}{d t}}=\frac{\sec ^{2} t}{\operatorname{cosec\: t}-\sin t} \\ &=\frac{\sec ^{2} t \cdot \sin t}{\cos ^{2} t}=\sec ^{2} t \tan t \\ &\left.\frac{d^{2} y}{d x^{2}}\right]_{-\frac{\pi}{4}}=2 \sqrt{2} \times 1 \\ &=2 \sqrt{2} \end{aligned}

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