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Higher-Order Derivatives

You’ll be able to

Differentiating again and again

The derivative f′(x) is itself a function, so you can differentiate it again to get the second derivative f″(x), and again for the third f‴(x), and so on. Notation: f″(x) or d²y/dx² for the second derivative. Each application just uses the same rules on the previous result — nothing new to learn, only careful repetition.

What the second derivative means

The second derivative is the rate of change of the rate of change. If f′ tells you how fast f is changing, f″ tells you how fast that slope is changing — which governs whether a graph curves upward or downward (concavity, coming in Unit 5). In motion problems the second derivative of position is acceleration, the rate at which velocity changes.

Position, velocity, acceleration

For an object with position s(t), the first derivative is velocity v(t) = s′(t) and the second derivative is acceleration a(t) = s″(t) = v′(t). This chain — position, then velocity, then acceleration — is one of the most tested ideas in the course, so keep the order straight: each derivative strips one layer off position.

Second derivative and motion
v(t) = s′(t) · a(t) = s″(t) = v′(t)
Velocity is the first derivative of position; acceleration is the second derivative of position (the derivative of velocity).
Worked example

A particle’s position is s(t) = t³ − 6t² + 9t. Find its velocity and acceleration functions, then the acceleration at t = 3.

  1. 1.Velocity is the first derivative: v(t) = s′(t) = 3t² − 12t + 9.
  2. 2.Acceleration is the second derivative: a(t) = v′(t) = 6t − 12.
  3. 3.Evaluate acceleration at t = 3: a(3) = 6(3) − 12 = 18 − 12.
  4. 4.That gives 6.
Answer: v(t) = 3t² − 12t + 9, a(t) = 6t − 12, and a(3) = 6 (units of distance per time squared).
Tip

Label each derivative as you go — s, then v = s′, then a = s″ — so you never lose track of which layer you are on. Miscounting derivatives (reporting velocity when acceleration was asked) is a needless lost point.

Checkpoint

What is the second derivative of f(x) = x⁴?

Checkpoint

If a particle has position s(t) = t² + 4t, what is its acceleration?

On the exam

On motion problems, read the question’s verb carefully: "how fast" usually means velocity (first derivative), while "how is the velocity changing" means acceleration (second derivative). Matching the derivative order to the phrasing is worth easy points.

Answer the 2 checkpoints as you read.

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