Method of DifferentiationeasyFree

Method of Differentiation: Second Derivative Single Valued Function Parametrically Repr

JEE Maths question with a full step-by-step solution.

Question
The second derivative of a single valued function parametrically represented by x=ϕ(t)x = \phi(t) and y=ψ(t)y = \psi(t), where α<t<β\alpha < t < \beta and ϕ(t)\phi(t), ψ(t)\psi(t) are differentiable functions, is given by:
Ad2ydx2=dxdtd2ydt2−d2xdt2dydt(dxdt)3\dfrac{d^2 y}{dx^2} = \dfrac{\frac{dx}{dt}\frac{d^2 y}{dt^2} - \frac{d^2 x}{dt^2}\frac{dy}{dt}}{\left(\frac{dx}{dt}\right)^3}correct
Bd2ydx2=dxdtd2ydt2−d2xdt2dydt(dxdt)2\dfrac{d^2 y}{dx^2} = \dfrac{\frac{dx}{dt}\frac{d^2 y}{dt^2} - \frac{d^2 x}{dt^2}\frac{dy}{dt}}{\left(\frac{dx}{dt}\right)^2}
Cd2ydx2=d2xdt2dydt−dxdtd2ydt2(dxdt)3\dfrac{d^2 y}{dx^2} = \dfrac{\frac{d^2 x}{dt^2}\frac{dy}{dt} - \frac{dx}{dt}\frac{d^2 y}{dt^2}}{\left(\frac{dx}{dt}\right)^3}
Dd2ydx2=d2xdt2dydt−d2ydt2dxdt(dydt)3\dfrac{d^2 y}{dx^2} = \dfrac{\frac{d^2 x}{dt^2}\frac{dy}{dt} - \frac{d^2 y}{dt^2}\frac{dx}{dt}}{\left(\frac{dy}{dt}\right)^3}
Solution
Step 1: First derivative
dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}
 Step 2: Differentiate with respect to xx via the chain rule
d2ydx2=ddt(dy/dtdx/dt)⋅dtdx=dxdtd2ydt2−dydtd2xdt2(dxdt)2⋅1dxdt\frac{d^2 y}{dx^2} = \frac{d}{dt}\left(\frac{dy/dt}{dx/dt}\right) \cdot \frac{dt}{dx} = \frac{\frac{dx}{dt}\frac{d^2 y}{dt^2} - \frac{dy}{dt}\frac{d^2 x}{dt^2}}{\left(\frac{dx}{dt}\right)^2} \cdot \frac{1}{\frac{dx}{dt}}
 Step 3: Combine
d2ydx2=dxdtd2ydt2−d2xdt2dydt(dxdt)3\frac{d^2 y}{dx^2} = \frac{\frac{dx}{dt}\frac{d^2 y}{dt^2} - \frac{d^2 x}{dt^2}\frac{dy}{dt}}{\left(\frac{dx}{dt}\right)^3}
 Answer: (1)
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