If $x=a \sin pt$, $y=b \cos pt$ find $\dfrac{d^2 y}{dx^2}$ at $t=0$
$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ $\cdots$ (1)
$\dfrac{dy}{dt}=-bp\sin \left(pt\right)$ $\cdots$ (2)
$\dfrac{dx}{dt}=ap \cos \left(pt\right)$ $\cdots$ (3)
$\therefore$ In view of equations (2) and (3) equation (1) becomes
$\dfrac{dy}{dx}=\dfrac{-bp \sin \left(pt\right)}{ap \cos \left(pt\right)}$
i.e. $\dfrac{dy}{dx}= \dfrac{-b}{a} \tan \left(pt\right)$
$\therefore$ $\dfrac{d^2 y}{dx^2}=\dfrac{d}{dx}\left[\dfrac{-b}{a} \tan \left(pt\right)\right]=\dfrac{-b}{a}\times p \times \sec^2 \left(pt\right) \times \dfrac{dt}{dx}$ $\cdots$ (4)
From equation (3), $\dfrac{dt}{dx}= \dfrac{1}{dx/dt}=\dfrac{1}{ap}\sec \left(pt\right)$ $\cdots$ (5)
Substituting the value of $\dfrac{dt}{dx}$ from equation (5) in equation (4) gives
$\dfrac{d^2 y}{dx^2} = \dfrac{-b}{a} \times p \times \sec^2 \left(pt\right)\times \dfrac{1}{ap}\times \sec \left(pt\right)$
i.e. $\dfrac{d^2 y}{dx^2} = \dfrac{-b}{a^2}\sec^3 \left(pt\right)$
$\therefore$ $\dfrac{d^2 y}{dx^2} \bigg|_{t=0} = \dfrac{-b}{a^2} \sec^3 \left(0\right) = \dfrac{-b}{a^2}$