If x = t^3, y = t^2 then (d^2y)/(d^2x) is equal to:

The correct answer is Option (4) → \(\displaystyle\frac{-2}{9t^4}\)

Here x and y are given in terms of a parameter t:

    \( \displaystyle \frac{dx}{dt} = 3t^2,\quad \frac{dy}{dt} = 2t \)

    \( \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{3t^2} = \frac{2}{3t} \)

For the second derivative, \(\frac{dy}{dx}\) is differentiated with respect to t and then divided by \(\frac{dx}{dt}\) (chain rule):

    \( \displaystyle \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{2}{3t}\right) \times \frac{dt}{dx} \)

    \( \displaystyle = -\frac{2}{3t^2} \times \frac{1}{3t^2} \)

    \( \displaystyle = \frac{-2}{9t^4} \)

Tip: \(\frac{d^2y}{dx^2}\) is not \(\frac{d^2y/dt^2}{d^2x/dt^2}\). Here that ratio would give \(\frac{2}{6t} = \frac{1}{3t}\), which is wrong.

Note: It seems that there is a mistake in the original NTA paper, which prints \(\frac{d^2y}{d^2x}\) when it should read \(\frac{d^2y}{dx^2}\). The question can still be solved because the second derivative of y with respect to x is clearly meant, and Option (4) is its value.