If the solve the problem
Question:

If $x=a(\theta-\sin \theta), y=a(1+\cos \theta)$ find $\frac{d^{2} y}{d x^{2}}$

Solution:

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,

$x=a(\theta-\sin \theta) \ldots \ldots$ equation 1

$y=a(1+\cos \theta) \ldots \ldots$ equation 2

to 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.

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\frac{\mathrm{d}}{\mathrm{d} \theta} \mathrm{a}(\theta-\sin \theta)=\mathrm{a}(1-\cos \theta) \ldots \ldots$ equation 3

Similarly,

$\frac{d y}{d \theta}=\frac{d}{d \theta} a(1+\cos \theta)=-a \sin \theta \ldots \ldots$ equation 4

$\left[\because \frac{d}{d x} \cos x=-\sin x, \frac{d}{d x} \sin x=\cos x\right.$

$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{-a \sin \theta}{a(1-\cos \theta)}=\frac{-\sin \theta}{(1-\cos \theta)} \ldots . . e q u a t i o n 5$

Differentiating again w.r.t $x$ :

$\frac{d}{d x}\left(\frac{d y}{d x}\right)=-\frac{d}{d x}\left(\frac{\sin \theta}{1-\cos \theta}\right)$

Using product rule and chain rule of differentiation together:

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\left\{-\frac{1}{1-\cos \theta} \frac{\mathrm{d}}{\mathrm{d} \theta} \sin \theta-\sin \theta \frac{\mathrm{d}}{\mathrm{d} \theta} \frac{1}{(1-\cos \theta)}\right\} \frac{\mathrm{d} \theta}{\mathrm{dx}}$

Apply chain rule to determine $\frac{\mathrm{d}}{\mathrm{d} \theta} \frac{1}{(1-\cos \theta)}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\left\{\frac{-\cos \theta}{1-\cos \theta}+\frac{\sin ^{2} \theta}{(1-\cos \theta)^{2}}\right\} \frac{1}{\mathrm{a}(1-\cos \theta)}$ [using equation 3]

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\left\{\frac{-\cos \theta(1-\cos \theta)+\sin ^{2} \theta}{(1-\cos \theta)^{2}}\right\} \frac{1}{\mathrm{a}(1-\cos \theta)}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\left\{\frac{-\cos \theta+\cos ^{2} \theta+\sin ^{2} \theta}{(1-\cos \theta)^{2}}\right\} \frac{1}{\mathrm{a}(1-\cos \theta)}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\left\{\frac{1-\cos \theta}{(1-\cos \theta)^{2}}\right\} \frac{1}{\mathrm{a}(1-\cos \theta)}\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right]$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{1}{\mathrm{a}(1-\cos \theta)^{2}}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{1}{\mathrm{a}\left(2 \sin ^{2} \frac{\theta}{2}\right)^{2}}\left[\because 1-\cos \theta=2 \sin ^{2} \theta / 2\right]$

$\therefore \frac{d^{2} y}{d x^{2}}=\frac{1}{4 a} \operatorname{cosec}^{4} \frac{\theta}{2}$