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  • Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 44 maths

Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 44 maths

Answers (1)

Answer:

\frac{1}{a \sin ^{2} t \cos t}

Hint:

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

Given:

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

Solution:

        \begin{aligned} &x=a \sin t \\ &\frac{d x}{d t}=a \cos t \\ &y=a\left(\cos t+\log \tan \frac{t}{2}\right) \\ &\frac{d y}{d t}=a\left(-\sin t+\cot \frac{t}{2} \times \sec ^{2} \frac{t}{2} \times \frac{t}{2}\right) \end{aligned}

        \begin{aligned} &=a\left(-\sin t+\frac{1}{2 \sin \left(\frac{t}{2}\right) \times \cos \left(\frac{t}{2}\right)}\right) \\ &\frac{d y}{d t}=a\left(-\sin t+\frac{1}{\sin t}\right) \\ &=a\left(\frac{-\sin ^{2} t+1}{\sin t}\right) \Rightarrow \frac{a \cos ^{2} t}{\sin t} \end{aligned}

        \begin{aligned} &\therefore \frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{\frac{a \cos ^{2} t}{\sin t}}{a \cos t} \\ &=\frac{\cos t}{\sin t}=\cot t \\ &\text { Again } \frac{d^{2} y}{d x^{2}}=-cosec^{2} t \frac{d t}{d x}=cose c^{2} t \times \frac{1}{a \cos t} \\ &=\frac{1}{a \sin ^{2} t \cos t} \end{aligned}

Posted by

Gurleen Kaur

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