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  • Explain solution RD Sharma class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question question 18

Explain solution RD Sharma class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question question 18

Answers (1)

Answer:

                (b)

Hint:

We must know about the derivative of trigonometric function.

Given:

                y=\frac{2}{\sqrt{a^{2}-b^{2}}} \tan ^{-1}\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right)

Explanation:

                y=\frac{2}{\sqrt{a^{2}-b^{2}}} \tan ^{-1}\left(\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right)

                \begin{aligned} &\\ &\frac{d y}{d x}=\frac{2}{\sqrt{a^{2}-b^{2}}} \cdot \frac{1}{1+\left(\frac{a-b}{a+b}\right) \tan ^{2}\left(\frac{x}{2}\right)} \times \sqrt{\frac{a-b}{a+b}} \frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right) \end{aligned}

                \frac{d y}{d x}=\frac{2}{\sqrt{a^{2}-b^{2}}} \cdot \frac{(a+b)}{(a+b)+(a-b) \tan ^{2}\left(\frac{x}{2}\right)} \times \sqrt{\frac{a-b}{a+b}} \frac{1}{2} \sec ^{2}\left(\frac{x}{2}\right)

                \frac{d y}{d x}=\sqrt{\frac{1}{a^{2}-b^{2}} \times \frac{a-b}{a+b} \times(a+b)^{2}} \times \frac{1+\tan ^{2}\left(\frac{x}{2}\right)}{a\left(1+\tan ^{2}\left(\frac{x}{2}\right)+b\left(1-\tan ^{2}\left(\frac{x}{2}\right)\right)\right)}

                \frac{d y}{d x}= \frac{1}{a+b\left(\frac{1-\tan ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}\right)}

                \begin{aligned} \\ \frac{d y}{d x}=& \frac{1}{a+b \cos x} \end{aligned}

Differentiate again,

                \frac{d^{2} y}{d x^{2}}=-\frac{-b \sin x}{(a+b \cos x)^{2}} \\  

                \begin{aligned} & &\frac{d^{2} y}{d x^{2}}=\frac{b \sin x}{(a+b \cos x)^{2}} \end{aligned}

 

Posted by

infoexpert27

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