If \(x = 4t^{2}, y = \displaystyle\frac{3}{t^{3}}\), then \(\displaystyle\frac{d^{2}y}{dx^{2}}\) at \(t = 1\) is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{45}{64}\)
For parametric functions, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\):
\( \displaystyle \frac{dx}{dt} = 8t, \quad \frac{dy}{dt} = -\frac{9}{t^{4}} \)
\( \displaystyle \frac{dy}{dx} = -\frac{9}{t^{4}} \times \frac{1}{8t} = -\frac{9}{8t^{5}} \)
The second derivative is the derivative of \(\frac{dy}{dx}\) with respect to t, divided again by \(\frac{dx}{dt}\):
\( \displaystyle \frac{d}{dt}\left(-\frac{9}{8t^{5}}\right) = \frac{45}{8t^{6}} \)
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \frac{45}{8t^{6}} \times \frac{1}{8t} = \frac{45}{64t^{7}} \)
At \(t = 1\):
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \frac{45}{64} \)