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If x=t2x = t^2 and y=tty = t\sqrt{t}, then d2ydx2\dfrac{d^2y}{dx^2} at t=14t = \dfrac{1}{4} is :

Solution

✅ Correct Option: 1

dydx=(3/2)t1/22t=34t−1/2\dfrac{dy}{dx}=\dfrac{(3/2)t^{1/2}}{2t}=\dfrac34 t^{-1/2}.

Differentiate w.r.t. tt: −38t−3/2-\dfrac38 t^{-3/2}, then divide by dxdt=2t\dfrac{dx}{dt}=2t to get −316t−5/2-\dfrac{3}{16}t^{-5/2}.

At t=14t=\dfrac14, t−5/2=32t^{-5/2}=32, so the value is −6-6.

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