If dxdt=3\displaystyle\frac{dx}{dt} = 3dtdx=3 and dydt=cos(2t)\displaystyle\frac{dy}{dt} = \cos(2t)dtdy=cos(2t), then d2ydx2=\displaystyle\frac{d^2y}{dx^2} =dx2d2y=
−2sin(2t)9\displaystyle\frac{-2\sin(2t)}{9}9−2sin(2t)
−2sin(2t)3\displaystyle\frac{-2\sin(2t)}{3}3−2sin(2t)
−2sin(2t)-2\sin(2t)−2sin(2t)
2sin(2t)9\displaystyle\frac{2\sin(2t)}{9}92sin(2t)
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