Use the chain rule to find the first and second derivatives of [tex]y[/tex] with respect to [tex]x[/tex].
[tex]\dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot\dfrac{dt}{dx} = \dfrac{dy/dt}{dx/dt}[/tex]
Differentiate the parametric equations with respect to [tex]t[/tex].
[tex]x(t) = t^2 \implies \dfrac{dx}{dt} = 2t[/tex]
[tex]y(t) = t^3 - 3t \implies \dfrac{dy}{dt} = 3t^2 - 3[/tex]
[tex]\implies \dfrac{dy}{dx} = \dfrac{3t^2 - 3}{2t}[/tex]
Let [tex]f(t) = \frac{dy}{dx}[/tex]. By the chain rule,
[tex]\dfrac{d^2y}{dx^2} = \dfrac{df}{dx} = \dfrac{df}{dt} \cdot \dfrac{dt}{dx} = \dfrac[df/dt}{dx/dt}[/tex]
Differentiate [tex]f[/tex].
[tex]f(t) = \dfrac{3t^2 - 3}{2t} = \dfrac32t - \dfrac3{2t} \implies \dfrac{df}{dt} = \dfrac{d^2y}{dx^2} = \dfrac32 + \dfrac3{2t^2}[/tex]
At [tex]t=2[/tex], the second derivative has a positive sign, so the curve is concave upward at this point.
