Topic 3.6 Calculating Higher-Order Derivatives: find second and higher-order derivatives and interpret their notation.

What this topic is asking

The College Board (Topic 3.6) introduces higher-order derivatives: the derivative of the derivative (the second derivative), the derivative of that (the third), and so on. You need to compute them by differentiating repeatedly and to read the notation - $f''(x)$, $\frac{d^2y}{dx^2}$, $y''$ - which all mean the same thing.

The notation

A reliable procedure

Computing higher-order derivatives is just the first-derivative process repeated. Differentiate to get $f'$, simplify, then differentiate that result to get $f''$, and continue as needed. Simplifying between steps keeps the next differentiation manageable.

What the second derivative means

The second derivative is not just an algebraic exercise; it carries two important meanings. Physically, if $s(t)$ is position, then $s'(t)$ is velocity (the rate at which position changes) and $s''(t)$ is acceleration (the rate at which velocity changes). A positive acceleration means the velocity is increasing; a negative acceleration means it is decreasing. This is why the particle-motion questions of Unit 4 lean directly on this topic. Geometrically, the second derivative measures concavity: $f''(x) > 0$ means the graph is concave up (curving like a cup, slopes increasing), and $f''(x) < 0$ means concave down (curving like a cap, slopes decreasing). A sign change in $f''$ marks an inflection point, where the curve switches concavity. These geometric meanings are developed fully in Unit 5, but they begin here, which is why the second derivative deserves attention beyond mere computation.

Implicit and repeated cases

Higher-order derivatives also arise on implicit curves: after finding $\frac{dy}{dx}$ implicitly, differentiating again (implicitly) gives $\frac{d^2 y}{dx^2}$, and you substitute the first derivative back in to express the answer in $x$ and $y$. For polynomials, repeated differentiation eventually reaches zero - the $(n+1)$th derivative of a degree-$n$ polynomial is $0$ - which is a useful check. For functions like $e^x$, every derivative is $e^x$ again; for $\sin x$, the derivatives cycle through $\cos x, -\sin x, -\cos x, \sin x$ with period four, so the $100$th derivative of $\sin x$ is $\sin x$ (since $100$ is a multiple of $4$). Recognizing these patterns lets you answer "find $f^{(n)}(x)$" questions without grinding through every step. On the AP exam, second derivatives are by far the most common, appearing in motion, concavity and the second-derivative test, so make $f''$ automatic before worrying about higher orders.