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If x=4t3x=4t^3 and y=6t2y=6t^2 then d2ydx2\frac{d^2y}{dx^2} at t=1t=1 is

Solution

✅ Correct Option: 2

dxdt=12t2\frac{dx}{dt}=12t^2 and dydt=12t\frac{dy}{dt}=12t, so dydx=1t\frac{dy}{dx}=\frac1t. Then d2ydx2=ddt(1t)÷dxdt=−1/t212t2=−112t4\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac1t\right)\div\frac{dx}{dt}=\frac{-1/t^2}{12t^2}=-\frac{1}{12t^4}, which is −112-\frac1{12} at t=1t=1.

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