#### Need solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 47

$\frac{(\sin 2 t+2 a t \cos 2 t)}{(\cos 2 t-2 a t \sin 2 t)}$

Hint:

You must know the derivative of cos and sin function

Given:

\begin{aligned} &x=a(\cos 2 t+2 t \sin 2 t) \\ &y=a(\sin 2 t-2 t \cos 2 t) \quad \text { find } \frac{d^{2} y}{d x^{2}} \end{aligned}

Solution:

\begin{aligned} &x=a(\cos 2 t+2 t \sin 2 t) \\ &\frac{d x}{d t}=a(-2 \sin 2 t+2 \sin 2 t+4 t \cos 2 t) \\ &=4 a t \cos 2 t \\ &\frac{d^{2} x}{d t^{2}}=4 a \cos 2 t-8 a t \sin 2 t \end{aligned}

\begin{aligned} &y=a(\sin 2 t-2 t \cos 2 t) \\ &\frac{d y}{d t}=a(2 \cos 2 t-2 \cos 2 t+4 t \sin 2 t) \\ &=4 a t \sin 2 t \\ &\frac{d^{2} y}{d t^{2}}=4 a \sin 2 t+8 a t \cos 2 t \end{aligned}

\begin{aligned} &\text { So, } \frac{d^{2} y}{d x^{2}}=\frac{4 \sin 2 t+8 a t \cos 2 t}{4 \cos 2 t-8 a t \sin 2 t} \\ &=\frac{4(\sin 2 t+2 a t \cos 2 t)}{4(\cos 2 t-2 a t \sin 2 t)} \\ &=\frac{(\sin 2 t+2 a t \cos 2 t)}{(\cos 2 t-2 a t \sin 2 t)} \end{aligned}