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If x=2t2+3,y=3t2+6t+5x = 2t^2 + 3, y = 3t^2 + 6t + 5, then the value of d2ydx2\frac{d^2y}{dx^2} is :

Solution

✅ Correct Option: 2

dxdt=4t\frac{dx}{dt}=4t, dydt=6t+6\frac{dy}{dt}=6t+6, so dydx=6t+64t=32+32t\frac{dy}{dx}=\frac{6t+6}{4t}=\frac{3}{2}+\frac{3}{2t}.

ddt(dydx)=−32t2\frac{d}{dt}\left(\frac{dy}{dx}\right)=-\frac{3}{2t^2}.

d2ydx2=−3/(2t2)4t=−38t3\frac{d^2y}{dx^2}=\frac{-3/(2t^2)}{4t}=-\frac{3}{8t^3}.

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