#### Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 8 maths

$\frac{-b^{4}}{a^{2}y^{3}}$

Hint:

You must know about derivative of

$tan\: \theta \: \: and\: \: sec\: \theta$

Given:

$I\! f\: \: x=a\: sec\: \theta ,y=b\: tan\: \theta$

Prove that

$\frac{d^{2}y}{dx^{2}}=\frac{-b^{4}}{a^{2}y^{3}}$

Solution:

$Let\: \: x=a\: sec\: \theta \\y=b\: tan\: \theta$

\begin{aligned} &\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right) \\ &x=a \sec \theta, y=b \tan \theta \end{aligned}

\begin{aligned} &\left.\frac{d x}{d \theta}=a \sec \theta \tan \theta \quad \text { ( } \frac{d \tan \theta}{d x}=\sec ^{2} x, \frac{d \sec \theta}{d x}=\sec \theta \tan \theta\right) \\ &\frac{d y}{d \theta}=b \sec ^{2} \theta \\ &\frac{d y}{d x}=\frac{b}{a} \frac{\sec \theta}{\tan \theta}=\frac{b}{a \sin \theta} \\ &\frac{d}{d \theta}\left(\frac{d y}{d x}\right)=-\frac{b}{a} \operatorname{cosec} \theta \cot \theta \end{aligned}

\begin{aligned} &=\frac{-\frac{b}{a} \operatorname{cosec} \theta \cot \theta}{a \sec \theta \tan \theta} \\ &\frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2} \tan ^{3} \theta} y=b \tan \theta \\ &\frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2}\left(\frac{y}{b}\right)^{3} \theta} y=\frac{-b^{4}}{a^{2} y^{3}} \end{aligned}