Let x = t^2,y=t^3 . Then (d^2y)/(dx^2) is equal to

The correct answer is Option (2) → \(\displaystyle\frac{3}{4t}\)

For a curve in parametric form:

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

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

The second derivative is taken with respect to x, so \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\) must be divided by dx/dt:

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

    \( \displaystyle \frac{d^2y}{dx^2} = \frac{3}{2} \cdot \frac{1}{2t} = \frac{3}{4t} \)

Tip: \(\frac{d^2y}{dx^2}\) is not just \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\), which here gives \(\frac{3}{2}\) (Option (1)), and not \(\frac{d^2y/dt^2}{d^2x/dt^2}\), which here gives \(\frac{6t}{2} = 3t\) (Option (3)).

NCERT Exemplar:
Continuity and Differentiability