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  • Need solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 43

Need solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 43

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}

Posted by

Gurleen Kaur

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