Lesson 15: Higher-Order Derivatives

(a) Opening Question

A ball is thrown straight up. Its height above the ground, in feet, after t seconds is

s(t) = -16t² + 64t

Without a calculator, answer these:

1. Find s'(t). What does it tell you physically?
2. Now take the derivative again to get s''(t). What is s''(t), and why is it a constant?

Take a moment before reading on. You already know how to differentiate -16t² + 64t: the power rule gives s'(t) = -32t + 64. That first derivative is the ball's velocity — how fast its height is changing, and in which direction. But there is no rule that says you can only differentiate once. Differentiate the velocity and you get s''(t) = -32. That constant is the acceleration — and -32 ft/s² is exactly the pull of gravity. The velocity is always changing at the same steady rate, which is why s'' is constant.

This lesson is about differentiating more than once: what it means, how to do it cleanly, and why it shows up everywhere on the AP exam.

(b) Core Concepts

Differentiating more than once

When you differentiate f(x), you get a new function f'(x). That function can be differentiated too. The result is the second derivative, written f''(x). Keep going and you get the third derivative f'''(x), and so on.

Definition. The second derivative of f is the derivative of the first derivative: f''(x) = d/dx[f'(x)]. More generally, the nth derivative f⁽ⁿ⁾(x) is the result of differentiating f a total of n times.

After three derivatives, the "prime" notation gets clumsy, so we switch to a parenthesized superscript: f⁽⁴⁾(x), f⁽⁵⁾(x), ..., f⁽ⁿ⁾(x). The parentheses matter: f⁴(x) would mean [f(x)]⁴ (the function raised to the fourth power), while f⁽⁴⁾(x) means the fourth derivative. These are completely different.

Notation summary. All of the following name the second derivative of y = f(x):

f''(x)      y''      d²y/dx²      d²/dx²[ f(x) ]      D²f

Read d²y/dx² as "d-squared y, d-x-squared." The placement of the 2's is a convention, not algebra: the top 2 sits on the d, the bottom 2 sits on the x. The third derivative is d³y/dx³, and the nth is dⁿy/dxⁿ.

Higher-order derivative of a polynomial

Each derivative of a polynomial lowers the degree by one, so eventually you hit zero. Take

f(x) = x⁵ - 4x³ + 2x² - 7x + 9

For a degree-n polynomial, the nth derivative is the constant n!·(leading coefficient) and every derivative after that is 0. Here f⁽⁵⁾(x) = 5!·1 = 120.

The sin/cos cycle

Trig derivatives repeat with period 4. Watch sine cycle:

f(x)    = sin x
f'(x)   = cos x
f''(x)  = -sin x
f'''(x) = -cos x
f⁽⁴⁾(x) = sin x      ← back to the start

So f⁽⁴⁾ = f, and the pattern restarts every four derivatives. To find a high-order derivative of sin x, divide the order by 4 and use the remainder:

remainder 0 →  sin x
remainder 1 →  cos x
remainder 2 → -sin x
remainder 3 → -cos x

Example: the 59th derivative of sin x. Since 59 = 4·14 + 3, the remainder is 3, so f⁽⁵⁹⁾(x) = -cos x. Cosine cycles the same way, just shifted (start it at -sin x).

The fixed point: eˣ

The exponential eˣ is its own derivative, so every derivative of eˣ is eˣ. We say eˣ is a fixed point of differentiation:

f(x) = eˣ → f'(x) = eˣ → f''(x) = eˣ → ... → f⁽ⁿ⁾(x) = eˣ

With a chain-rule inner function, a constant factor compounds: if f(x) = e^{kx}, then f⁽ⁿ⁾(x) = kⁿ e^{kx}. For example d³/dx³[e^{2x}] = 2³e^{2x} = 8e^{2x}. Logarithms also fall into a pattern: from f(x) = ln x you get f'(x) = 1/x, f''(x) = -1/x², f'''(x) = 2/x³, f⁽⁴⁾(x) = -6/x⁴.

Position → Velocity → Acceleration

This is the headline application and a recurring AP theme.

If s(t) gives the position of an object along a line at time t, then

- velocity is the first derivative: v(t) = s'(t)

- acceleration is the second derivative: a(t) = v'(t) = s''(t)

Velocity is the rate of change of position; acceleration is the rate of change of velocity. The sign of velocity tells you direction (positive = moving in the positive direction, negative = moving in the negative direction). The object is speeding up when velocity and acceleration have the same sign, and slowing down when they have opposite signs. (Memorize that — it's a classic AP justification.)

Higher-order implicit differentiation (brief)

When a curve is given implicitly, you can still find d²y/dx². Differentiate the equation, solve for dy/dx, then differentiate again — and every time you hit a y, remember y is a function of x, so chain-rule it. Take the circle x² + y² = 25. Differentiating once:

2x + 2y·(dy/dx) = 0   ⟹   dy/dx = -x/y

Differentiate -x/y using the quotient rule, treating y as a function of x:

d²y/dx² = -[ (1)(y) - x(dy/dx) ] / y²

Substitute dy/dx = -x/y:

d²y/dx² = -[ y - x(-x/y) ] / y² = -[ y + x²/y ] / y² = -(y² + x²)/y³

Finally use the original equation x² + y² = 25:

d²y/dx² = -25/y³

The key move: after solving for dy/dx, substitute it back in so the final answer has no dy/dx left in it.

(c) Worked Examples

Example 1 — nth derivative of a polynomial (NO CALC)

Problem. Let f(x) = 2x⁴ - 5x² + 8x - 1. Find f''(x), f⁽⁴⁾(x), and f⁽⁵⁾(x).

Strategy. Apply the power rule repeatedly. The degree is 4, so the 4th derivative is a constant and the 5th is 0.

Solution.

f'(x)   = 8x³ - 10x + 8
f''(x)  = 24x² - 10
f'''(x) = 48x
f⁽⁴⁾(x) = 48
f⁽⁵⁾(x) = 0

Justification. f is a degree-4 polynomial, so f⁽⁴⁾(x) = 4!·2 = 48 (a constant), and every derivative beyond the 4th is 0.

Example 2 — The sin/cos cycle (NO CALC)

Problem. Find f⁽¹⁰⁰⁾(x) for f(x) = sin x, and g⁽⁵⁰⁾(x) for g(x) = cos x.

Strategy. Use period-4 cycling: divide the order by 4 and read off the remainder.

Solution.

• 100 = 4·25 + 0, remainder 0. Starting from sin x, remainder 0 returns to the start: f⁽¹⁰⁰⁾(x) = sin x.
• For cosine, the cycle is cos x → -sin x → -cos x → sin x → cos x (period 4). 50 = 4·12 + 2, remainder 2, which lands two steps in: g⁽⁵⁰⁾(x) = -cos x.

Justification. Derivatives of sin x and cos x repeat every 4 steps, so only the order mod 4 matters.

Example 3 — Position, velocity, acceleration (NO CALC)

Problem. A particle moves along a line with position s(t) = t³ - 6t² + 9t + 2 for t ≥ 0 (s in meters, t in seconds).

(i) Find v(t) and a(t). (ii) Find the velocity and acceleration at t = 2. (iii) Is the particle speeding up or slowing down at t = 2? Justify.

Strategy. Differentiate once for velocity, twice for acceleration. Compare signs to decide speeding up vs. slowing down.

Solution.

v(t) = s'(t)  = 3t² - 12t + 9
a(t) = s''(t) = 6t - 12

At t = 2: v(2) = 3(4) - 24 + 9 = -3 m/s and a(2) = 12 - 12 = 0 m/s².

Hmm — a(2) = 0, so the speeding-up/slowing-down test is inconclusive at exactly t = 2 (acceleration is momentarily zero). Let's instead answer at t = 1: v(1) = 3 - 12 + 9 = 0, also a turning point. Try t = 0.5: v(0.5) = 0.75 - 6 + 9 = 3.75 > 0 and a(0.5) = 3 - 12 = -9 < 0.

Justification. At t = 0.5, velocity is positive and acceleration is negative — opposite signs — so the particle is slowing down at t = 0.5. (At t = 2, v = -3 < 0 and a = 0, so the speed test is inconclusive there because acceleration is momentarily zero.)

Example 4 — Implicit second derivative (NO CALC)

Problem. For the curve y² = x³, find d²y/dx² in terms of x and y.

Strategy. Differentiate implicitly to get dy/dx, then differentiate again and substitute dy/dx back in.

Solution. Differentiate y² = x³:

2y·(dy/dx) = 3x²   ⟹   dy/dx = 3x²/(2y)

Differentiate dy/dx = 3x²/(2y) with the quotient rule (and d/dx[2y] = 2·dy/dx):

d²y/dx² = [ (6x)(2y) - 3x²·(2·dy/dx) ] / (2y)²
        = [ 12xy - 6x²·(dy/dx) ] / (4y²)

Substitute dy/dx = 3x²/(2y):

d²y/dx² = [ 12xy - 6x²·(3x²/(2y)) ] / (4y²)
        = [ 12xy - 9x⁴/y ] / (4y²)
        = 3x/y - 9x⁴/(4y³)

Justification. After two implicit differentiations, every dy/dx was replaced by its value, leaving an answer in x and y only. (Verified with sympy: d²y/dx² = -9x⁴/(4y³) + 3x/y.)

(d) Common Mistakes

Confusing f''(x) with [f'(x)]². f''(x) is "differentiate twice." [f'(x)]² is "differentiate once, then square." For f(x) = x²: f''(x) = 2, but [f'(x)]² = (2x)² = 4x². These are not the same. Fix: read f'' as a count of derivatives, never as an exponent.

Stopping one derivative too early. Acceleration is the second derivative of position. Students compute v(t) = s'(t) and then report it as acceleration. Fix: a = s''. If the question says "acceleration," differentiate twice from position.

Sign errors in the trig cycle. Going sin x → cos x → -sin x → -cos x → sin x, students drop a negative or restart the cycle wrong. Fix: write out all four steps once, then use order mod 4. Remainder 2 always means the negative of the original function.

Forgetting the chain rule on the second implicit derivative. When differentiating dy/dx again, terms containing y need a dy/dx factor (and you must substitute it back in). Treating y as a constant the second time is the most common implicit error. Fix: every d/dx of a y-term produces a dy/dx; substitute its value at the end.

Misreading the d²y/dx² notation. It is not (dy/dx)² and not d²y/d²x. The exponents sit on the d (top) and the variable x (bottom). Fix: memorize the placement — it names "second derivative," nothing is being squared.

(e) Practice Problems

(Short answer) Let s(t) = -16t² + 48t + 5 be the height (ft) of a ball at time t (s). Find v(t) and a(t), and state the value and physical meaning of a(t).

(Short answer) For the curve x² + y² = 25, you are told dy/dx = -x/y. Find d²y/dx² in terms of x and y. Show your substitution step.

(Justification) A particle has v(t) = t² - 4t + 3 and a(t) = 2t - 4. At t = 1, determine whether the particle is speeding up or slowing down. Justify your answer using the signs of velocity and acceleration.

(Interpretation) An object's velocity is v(t) = nDeriv of position s(t) = t³ - 6t². At t = 3, you compute v(3) = -9 and (by differentiating by hand) a(3) = 6. Explain the meaning of these two values in the context of the object's motion, and state whether it is speeding up or slowing down. (Note: the TI-84 nDeriv command gives only the first derivative; you must find a(t) by hand.)

(Justification) Explain why every derivative of f(x) = eˣ is eˣ, and use this to state f⁽¹⁰⁰⁾(x) and f⁽¹⁰⁰⁾(0).

(f) AP Exam Focus

Free-Response Question (NO CALCULATOR — Section II Part B style) · 9 points total

A particle moves along the x-axis. For time t ≥ 0, its position is given by

s(t) = t³ - 6t² + 9t + 1

where s is measured in meters and t in seconds.

(a) Find the velocity v(t) and the acceleration a(t). (2 points)

(b) Find all times t in the interval 0 ≤ t ≤ 4 at which the particle is at rest. (2 points)

(c) At t = 4, is the speed of the particle increasing or decreasing? Justify your answer. (3 points)

(d) Find the acceleration of the particle at the moment(s) when its velocity is zero, and interpret the meaning of the acceleration at the later such time. (2 points)

Model Solution

(a) Differentiate the position function once for velocity, twice for acceleration:

v(t) = s'(t)  = 3t² - 12t + 9
a(t) = s''(t) = 6t - 12

(1 pt v(t), 1 pt a(t).)

(b) The particle is at rest when v(t) = 0:

3t² - 12t + 9 = 0  ⟹  3(t² - 4t + 3) = 0  ⟹  3(t - 1)(t - 3) = 0

So t = 1 and t = 3, both in [0, 4]. (1 pt setting v = 0, 1 pt both answers.)

(c) At t = 4:

v(4) = 3(16) - 48 + 9 = 9 > 0
a(4) = 6(4) - 12 = 12 > 0

Since v(4) and a(4) are both positive (same sign), the speed of the particle is increasing at t = 4.

By the speeding-up criterion, the particle's speed is increasing because v(4) > 0 and a(4) > 0 have the same sign.

(1 pt v(4) = 9, 1 pt a(4) = 12, 1 pt correct conclusion with same-sign justification.)

(d) From part (b), velocity is zero at t = 1 and t = 3:

a(1) = 6(1) - 12 = -6 m/s²
a(3) = 6(3) - 12 = 6 m/s²

At the later time t = 3, the acceleration is 6 m/s². Interpretation: at t = 3 the particle is momentarily at rest, and since acceleration is positive, its velocity is increasing through zero — the particle is reversing direction from moving in the negative direction to moving in the positive direction. (1 pt both acceleration values, 1 pt interpretation.)

Scoring Commentary — where students lose points

• Part (a): Reporting v(t) as the acceleration (stopping one derivative early) loses the acceleration point. Acceleration is s'', the second derivative.
• Part (b): Forgetting that t = 3 is also in [0, 4] and giving only t = 1 loses a point. Always check all solutions against the interval.
• Part (c): Writing only "speed is increasing" with no reason earns the computation points but loses the justification point. The AP rubric requires citing that v and a have the same sign. Saying "acceleration is positive so it speeds up" is incomplete — a particle moving in the negative direction with positive acceleration is slowing down. You must compare signs.
• Part (d): A bare number with no sentence of interpretation loses the interpretation point. "Interpret" means explain the physical meaning in context (rest, direction reversal), not just compute.

d²y/dx² = -[ (1)(y) - x(dy/dx) ] / y²

= -[ y - x(-x/y) ] / y² = -[ y + x²/y ] / y² = -(y² + x²)/y³ = -25/y³