#### Provide Solution for RD Sharma Class 12 Chapter 11 Higher Order Derivatives Exercise Fill in the blanks Question 2

Answer:  $\frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2}} \sec ^{3} \theta$

Hint: Differentiate x & y w.r.t  $\theta$. Use   $\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$.   Again, differentiate w.r.t x.

Given: $y=b \cos \theta \& \quad x=a \sin \theta$

Solution:

It is given that: $x=a \sin \theta \& y=b \cos \theta$

Differentiating y w.r.t t,

$\frac{d y}{d \theta}=-b \sin \theta$                                                                               …(i)

Differentiating y w.r.t t,

$\frac{d x}{d \theta}=a \cos \theta$                                                                   …(ii)

So, $\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$

$\frac{d y}{d x}=\frac{-b \sin \theta}{a \cos \theta}$                        …(iii)

Again, differentiate (iii) w.r.t x:

\begin{aligned} &\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(\frac{-b \sin \theta}{a \cos \theta}\right) \\ & \end{aligned}

$\frac{d^{2} y}{d x^{2}}=\frac{d}{d \theta}\left(\frac{-b \sin \theta}{a \cos \theta}\right) \cdot \frac{d \theta}{d x}$

\begin{aligned} &=\frac{d}{d \theta}\left(\frac{-b}{a} \tan \theta\right) \cdot \frac{1}{a \cos \theta} \\ & \end{aligned}                                      [ using (ii) ]

$=\frac{-b}{a} \sec ^{2} \theta \cdot\left(\frac{1}{a \cos \theta}\right) \\$

$=\frac{-b}{a^{2}} \sec ^{3} \theta$

Thus,  $\frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2}} \sec ^{3} \theta$.