Second Derivatives of Parametric Equations BC only

$\frac{d^{2}y}{dx^{2}}$ is the derivative of $\frac{dy}{dx}$ with respect to $x$, and since only $t$-derivatives are available the chain rule converts: $\frac{d^{2}y}{dx^{2}} = \frac{d}{dt}\!\left(\frac{dy}{dx}\right) \div \frac{dx}{dt}$. Read as two operations, it says differentiate the slope with respect to $t$, then convert that rate in $t$ into a rate in $x$ : the same conversion 9.1 applied to $y$, applied again to the slope, which is why the second derivative needs two divisions. For $x = t^{2}+1$, $y = t^{3}-3t$ the result is $\frac{3(t^{2}+1)}{4t^{3}}$, which is $\frac{15}{32}$ at $t = 2$, while stopping one step early gives $\frac{15}{8}$.

The tempting $\frac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}}$ is not the second derivative and is not any derivative: it gives $6$ here against $\frac{15}{32}$, and it fails because the chain rule was never a rule about matching orders. Differentiating $\frac{dy}{dx}\cdot\frac{dx}{dt}$ a second time produces two terms and the naive quotient discards one. Concavity is read from the sign as in Unit 5, and here the numerator is always positive so the sign is the sign of $t$: concave down for $t < 0$, up for $t > 0$. Note that the sign of $t$ is not the sign of $x$, and that the flip at $t = 0$ is a vertical tangent rather than an inflection point.