If \(x=t^4,\ y=t\), then \(\displaystyle\frac{d^2y}{dx^2}\) is given by ________.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-\displaystyle\frac{3}{16t^7}\)
For a curve in parametric form:
\( \displaystyle \frac{dx}{dt} = 4t^3, \qquad \frac{dy}{dt} = 1 \)
\( \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1}{4t^3} \)
The second derivative is the t-derivative of dy/dx, divided again by dx/dt:
\( \displaystyle \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{1}{4t^3}\right)\cdot\frac{dt}{dx} \)
\( \displaystyle = \left(-\frac{3}{4t^4}\right)\cdot\frac{1}{4t^3} \)
\( \displaystyle = -\frac{3}{16t^7} \)
Tip: \(\frac{d^2y}{dx^2}\) is not \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\). Forgetting to multiply by \(\frac{dt}{dx}\) gives \(-\frac{3}{4t^4}\), Option (3).