#### Please solve RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 17 maths textbook solution

\begin{aligned} &\text { } 3 \sin ^{2} \theta\left[5 \cos ^{2} \theta-1\right]\\ \end{aligned}

Hint:

\begin{aligned} &\text { You must know about derivative of } \cos \theta\: \&\: \sin \theta\\ \end{aligned}

Given:

\begin{aligned} &\text { If } x=\cos \theta, y=\sin ^{3} \theta\\ &\text { Prove that } y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=3 \sin ^{2} \theta\left[5 \cos ^{2} \theta-1\right] \end{aligned}

Solution:

\begin{aligned} &\text { Let } x=\cos \theta, y=\sin ^{3} \theta\\ \end{aligned}

\begin{aligned} &\frac{d x}{d \theta}=-\sin \theta \\ &\frac{d y}{d \theta}=3 \sin ^{2} \theta \cos \theta \\ &\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 \sin ^{2} \theta \cos \theta}{-\sin \theta}=3 \sin \theta \cos \theta \end{aligned}

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

\begin{aligned} &y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=\frac{\sin ^{3} \theta\left(3 \cos ^{2} \theta-3 \sin ^{2} \theta\right)}{\sin \theta} \\ &y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=3 \sin ^{2} \theta\left[5 \cos ^{2} \theta-1\right] \end{aligned}

Hence proved