#### Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 49 maths textbook solution

$\frac{-(x^{2}+y^{2})}{y^{3}}$

Hint:

You must know the derivative of cos t and sin t function

Given:

\begin{aligned} &x=a \sin t-b \cos t \\ &y=a \cos t-b \sin t \quad \text { Prove } \frac{d^{2} y}{d x^{2}}=\frac{-\left(x^{2}+y^{2}\right)}{y^{3}} \end{aligned}

Solution:

\begin{aligned} &\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{a \sin t+b \cos t}{a \cos t+b \sin t}\\ &\text { Use quotient rule }\\ &\frac{U}{V}=\frac{U V^{\prime}-U^{\prime} V}{V^{2}} \end{aligned}

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{\frac{d\left(\frac{d y}{d x}\right)}{d t}}{\frac{d x}{d t}}\\ &=\frac{(-a \cos t-b \sin t)(a \cos t+b \sin t)-(a \sin t+b \cos t)(-a \sin t+b \cos t)}{(a \cos t+b \sin t)^{2}(a \cos t+b \sin t)}\\ &=\frac{(-a \cos t+b \sin t)^{2}-(b \cos t-a \sin t)^{2}}{y^{3}}\\ &=\frac{-y^{2}-x^{2}}{y^{3}}=\frac{-\left(x^{2}+y^{2}\right)}{y^{3}} \end{aligned}

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