If \( x=t^3,y=t^4 \), then \( \displaystyle\frac{d^2y}{dx^2} \) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{4}{9t^2} \)
\( \displaystyle \frac{dx}{dt}=3t^2 \)
\( \displaystyle \frac{dy}{dt}=4t^3 \)
\( \displaystyle \frac{dy}{dx}=\frac{4t^3}{3t^2}=\frac{4t}{3} \)
The second derivative is with respect to x, so \( \frac{dy}{dx} \) is differentiated with respect to t and divided by \( \frac{dx}{dt} \) again:
\( \displaystyle \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{4t}{3}\right)}{\frac{dx}{dt}} \)
\( \displaystyle =\frac{\frac{4}{3}}{3t^2} \)
\( \displaystyle =\frac{4}{9t^2} \)
Shortcut: Eliminate t: \( t=x^{1/3} \), so \( y=x^{4/3} \), \( \frac{dy}{dx}=\frac{4}{3}x^{1/3} \) and \( \frac{d^2y}{dx^2}=\frac{4}{9}x^{-2/3}=\frac{4}{9t^2} \).
Why the other options are incorrect
Option (1) \( \displaystyle\frac{4}{3} \) — Incorrect. \( \frac{4}{3} \) is only \( \frac{d}{dt}\left(\frac{dy}{dx}\right) \); it still has to be divided by \( \frac{dx}{dt}=3t^2 \).
Option (2) \( \displaystyle\frac{4}{3t} \) — Incorrect. At t = 1 this gives \( \frac{4}{3} \), while the second derivative there is \( \frac{4}{9} \).
Option (3) \( \displaystyle\frac{4}{3t^2} \) — Incorrect. This divides \( \frac{4}{3} \) by \( t^2 \) instead of by \( 3t^2 \).
Tip: In parametric form, \( \frac{d^2y}{dx^2}\ne\frac{d^2y/dt^2}{d^2x/dt^2} \); always divide \( \frac{d}{dt}\left(\frac{dy}{dx}\right) \) by \( \frac{dx}{dt} \).