Differentiating again AB & BC

Notation first. The second derivative is written $f''(x)$, or $\frac{d^2y}{dx^2}$, or $y''$. It measures how the rate is changing, which is what concavity and acceleration are.

Mechanically, the instruction is simple: differentiate, then differentiate the result. The trap is that the result is often more complicated than the original, so the second pass may need more rules than the first.

Take $y = \sin(2x)$. The first derivative is $2\cos(2x)$. The second is not $-\sin(2x)$; the chain rule applies again, and the constant 2 is still out front:

$$y'' = 2\cdot\left(-\sin(2x)\right)\cdot 2 = -4\sin(2x).$$

Each pass through a composition contributes its own inner factor, so an $n$th derivative of $\sin(kx)$ carries $k^n$. Losing that on the second pass is the signature error of this topic.

Differentiating with the power rule can turn a simple function into a product, and then the second pass needs the product rule even though the first did not. For $y = x^2e^x$:

$$y' = 2xe^x + x^2e^x.$$

Now $y'$ is a sum of two products, so the second derivative takes the product rule twice:

$$y'' = \left(2e^x + 2xe^x\right) + \left(2xe^x + x^2e^x\right) = e^x\left(x^2 + 4x + 2\right).$$

A reliable habit: after computing $y'$, look at it as a brand new problem and diagnose its structure from scratch, exactly as Topic 3.5 asks. Reusing the diagnosis from the first pass is what drops rules.

For the circle $x^2 + y^2 = 25$ we found $\frac{dy}{dx} = -\frac{x}{y}$. To get $\frac{d^2y}{dx^2}$, differentiate that expression with the quotient rule, remembering that $y$ is still a function of $x$:

$$\frac{d^2y}{dx^2} = -\frac{(1)(y) - (x)\frac{dy}{dx}}{y^2}.$$

That answer still contains $\frac{dy}{dx}$, so the last step is to substitute the first derivative back in:

$$\frac{d^2y}{dx^2} = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3} = -\frac{25}{y^3}.$$

Two steps get skipped. Differentiating $\frac{dy}{dx}$ as if $y$ were a constant loses the chain factor. And stopping before the substitution leaves an answer written in terms of an unknown, which on the exam is an incomplete response.

For position $s(t)$, the first derivative is velocity and the second is acceleration. A negative acceleration does not mean slowing down; it means velocity is decreasing, which speeds an object up if it is already moving in the negative direction. Speed increases exactly when velocity and acceleration share a sign.

For a graph, $f'' > 0$ means concave up and $f'' < 0$ means concave down. The circle result above is a good check on that: $-\frac{25}{y^3}$ is negative on the upper half where $y > 0$, and the top of a circle is indeed concave down. When a second derivative comes out in a form you can sanity-check against a picture, check it.