Method of DifferentiationeasyFree

Method of Differentiation — JEE Maths practice question

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

Question

The second derivative of a single valued function parametrically represented by

x = \phi(t)

and

y = \psi(t)

, where

\alpha < t < \beta

and

\phi(t)

\psi(t)

are differentiable functions, is given by:

\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

\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}

\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}

\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

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

Step 2: Differentiate with respect to

x

via the chain rule

\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

\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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