Divide by $dx/dt$ a second time BC only

$\frac{d^{2}y}{dx^{2}}$ means the derivative of $\frac{dy}{dx}$ with respect to $x$. We can only differentiate with respect to $t$, so the chain rule converts:

$$\frac{d^{2}y}{dx^{2}} = \frac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}.$$

Read it as two operations rather than one formula. Differentiate the slope with respect to $t$, which gives a rate in $t$. Then convert that rate into a rate in $x$, which is what dividing by $\frac{dx}{dt}$ does. It is the same conversion 9.1 performed on $y$, applied a second time to a different quantity.

That reading also explains the shape. Every time you want a rate with respect to $x$ and only have rates with respect to $t$, you divide by $\frac{dx}{dt}$. The first derivative needed one conversion; the second derivative needs one for $y$ and one for the slope, so two in total.

The question that catches the error, every time: after writing something down, ask what variable it is a rate in. $\frac{d}{dt}\!\left(\frac{dy}{dx}\right)$ is a rate in $t$. The problem asked for a rate in $x$. They are not the same number.

Take 9.1's curve, $x = t^{2} + 1$ and $y = t^{3} - 3t$, where $\frac{dy}{dx} = \frac{3t^{2} - 3}{2t}$.

1. Differentiate the slope with respect to $t$. By the quotient rule, $\frac{d}{dt}\!\left(\frac{3t^{2}-3}{2t}\right) = \frac{6t\cdot 2t - (3t^{2}-3)\cdot 2}{4t^{2}} = \frac{3(t^{2}+1)}{2t^{2}}$.
2. Divide by $\frac{dx}{dt} = 2t$. $\frac{d^{2}y}{dx^{2}} = \frac{3(t^{2}+1)}{2t^{2}} \cdot \frac{1}{2t} = \frac{3(t^{2}+1)}{4t^{3}}$.

At $t = 2$ that is $\frac{3(5)}{4(8)} = \frac{15}{32}$. Stopping after step 1 gives $\frac{15}{8}$, which is four times too large and looks entirely respectable.

A simplification worth knowing, because it halves the work: $\frac{3t^{2}-3}{2t} = \frac{3}{2}t - \frac{3}{2}t^{-1}$, whose $t$-derivative is $\frac{3}{2} + \frac{3}{2}t^{-2}$ by the power rule alone. Splitting the quotient before differentiating avoids the quotient rule entirely, and it is usually available.

The most tempting wrong answer is

$$\frac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}},$$

because it looks like 9.1's rule with every symbol upgraded. It is not the second derivative, and it is not any derivative at all. For the running example it gives $\frac{6t}{2} = 3t$, so $6$ at $t = 2$, against a true $\frac{15}{32}$. Those are not close, and no interval of $t$ makes them agree.

The reason the pattern fails is that the chain rule was never a rule about "matching orders". $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ came from solving $\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$, and there is no corresponding identity relating $\frac{d^{2}y}{dt^{2}}$ to $\frac{d^{2}y}{dx^{2}}$ by a single factor. Differentiating the product $\frac{dy}{dx}\cdot\frac{dx}{dt}$ a second time produces two terms, not one, and the extra term is exactly what the naive quotient discards.

Two smaller variants of the same failure:

1. Multiplying by $\frac{dx}{dt}$ instead of dividing, which gives $\frac{15}{2}$ here. The conversion always divides, in both derivatives.
2. Carrying an inverted first derivative into step 1. If $\frac{dy}{dx}$ was built as $\frac{dx/dt}{dy/dt}$, everything downstream is wrong and the second derivative will still simplify cleanly. That is CA1 propagating, and it is why 9.1's quotient is worth getting right before starting here.

The second derivative is used for the same thing here as in Unit 5: concavity, and it is read the same way. Positive means concave up, negative means concave down, both as seen in the $xy$-plane.

For the running example, $\frac{d^{2}y}{dx^{2}} = \frac{3(t^{2}+1)}{4t^{3}}$. The numerator is positive for every $t$, so the sign is entirely the sign of $t^{3}$, which is the sign of $t$. The curve is concave down for $t < 0$ and concave up for $t > 0$, which matches the picture: the upper branch of the loop arches over and the lower branch cups under.

Two cautions specific to the parametric setting:

1. The sign of $t$ is not the sign of $x$. Here $x = t^{2}+1$ takes the same value at $t$ and $-t$, so a single $x$ can sit on a concave-up branch and a concave-down branch at once. Concavity belongs to the branch, not to the $x$-value.
2. A sign change in $\frac{d^{2}y}{dx^{2}}$ need not be an inflection point. Here the sign flips at $t = 0$, which is where $\frac{dx}{dt} = 0$ and the second derivative is undefined rather than zero. That is the vertical tangent from 9.1, not an inflection.