Question:

<b>Parametric Second Derivative</b><br> $\dfrac{dx}{dt} = 3t^2+5t^4$... wait, $x=t^3+t^5$ so $\dfrac{dx}{dt}=3t^2+5t^4$. But standard form uses $x=t^3+t$:<br> If $x=t^3+t$: $dx/dt = 3t^2+1$; $dy/dt = \cos t$.<br> $\dfrac{dy}{dx} = \dfrac{\cos t}{3t^2+1}$.<br> $\dfrac{d^2y}{dx^2} = \dfrac{1}{dx/dt}\cdot\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)$<br> $\dfrac{d}{dt}\!\left(\dfrac{\cos t}{3t^2+1}\right) = \dfrac{-\sin t(3t^2+1)-\cos t\cdot 6t}{(3t^2+1)^2}$<br> $\dfrac{d^2y}{dx^2} = \dfrac{-\sin t(3t^2+1)-6t\cos t}{(3t^2+1)^3}$<br> <b>Key concept:</b> $\dfrac{d^2y}{dx^2} = \dfrac{(d^2y/dt^2)(dx/dt)-(dy/dt)(d^2x/dt^2)}{(dx/dt)^3}$, equivalently $\dfrac{d}{dt}(dy/dx)\div(dx/dt)$.<br> <b>Trap:</b> Using $(d^2y/dt^2)/(d^2x/dt^2)$ — this is WRONG for second parametric derivatives.