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

(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}