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The idea of parametric form of differentiation:

If $y=f(\theta)$ and $x=g(\theta)$, i.e. $y$ is a function of $\theta$ and $x$ is also some other function of $\theta$.

Then $d y / d \theta=f^{\prime}(\theta)$ and $d x / d \theta=g^{\prime}(\theta)$

We can write : $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}$

Given,

$y=\sin ^{3} \theta \ldots \ldots$ equation 1

$x=\cos \theta \ldots \ldots$ equation 2

To prove: $y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=3 \sin ^{2} \theta\left(5 \cos ^{2} \theta-1\right)$

We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.

Let's find $\frac{d^{2} y}{d x^{2}}$

$\mathrm{As}, \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

So, lets first find $\mathrm{dy} / \mathrm{dx}$ using parametric form and differentiate it again.

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=-\sin \theta \ldots \ldots \ldots \ldots$ equation 3

Applying chain rule to differentiate $\sin ^{3} \theta$ :

$\frac{d y}{d \theta}=3 \sin ^{2} \theta \cos \theta \ldots \ldots \ldots \ldots . .$ equation 4

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}=\frac{3 \sin ^{2} \theta \cos \theta}{-\sin \theta}=-3 \sin \theta \cos \theta \ldots \ldots \ldots \ldots$ equation 5

Again differentiating w.r.t $\mathrm{x}$ :

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}(-3 \sin \theta \cos \theta)$

Applying product rule and chain rule to differentiate:

$\frac{d^{2} y}{d x^{2}}=-3\left\{\sin \theta \frac{d}{d \theta} \cos \theta+\cos \theta \frac{d}{d \theta} \sin \theta\right\} \frac{d \theta}{d x}$

$\frac{d^{2} y}{d x^{2}}=3\left\{-\sin ^{2} \theta+\cos ^{2} \theta\right\} \frac{1}{\sin \theta}$

[using equation 3 to put the value of $d \theta / d x$ ]

Multiplying y both sides to approach towards the expression we want to prove-

$y \frac{d^{2} y}{d x^{2}}=3\left\{-\sin ^{2} \theta+\cos ^{2} \theta\right\} \frac{y}{\sin \theta}$

$y \frac{d^{2} y}{d x^{2}}=3\left\{-\sin ^{2} \theta+\cos ^{2} \theta\right\} \sin ^{2} \theta$

[from equation 1, substituting for $y$ ]

Adding equation 5 after squaring it:

$y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=3\left\{-\sin ^{2} \theta+\cos ^{2} \theta\right\} \sin ^{2} \theta+9 \sin ^{2} \theta \cos ^{2} \theta$

$y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=3 \sin ^{2} \theta\left\{-\sin ^{2} \theta+\cos ^{2} \theta+3 \cos ^{2} \theta\right\}$

$\mathrm{y} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}=3 \sin ^{2} \theta\left\{5 \cos ^{2} \theta-1\right\} \ldots \ldots .$ proved