Higher-Order Derivatives

Sit in a car on a smooth motorway doing a steady 70 mph and close your eyes. Can you tell how fast you are going? You can't. Steady speed feels exactly like sitting still — which is why you can pour a drink on a cruising aeroplane at 550 mph. Your body has no speedometer.

Now the driver puts their foot down. That you feel instantly — the seat pressing into your back. Your body is an accelerometer: it senses not position, not velocity, but the rate of change of velocity. And when the driver stamps on the brake, you feel something sharper still — the lurch, the sudden change in the push. That's the rate of change of acceleration.

Position, velocity, acceleration, lurch: each one is the derivative of the one before. Your body physically feels the second and third derivatives of your position. This page is about what "the derivative of the derivative" means, how to compute it, and why anyone — from a physicist to a roller-coaster engineer — would want to.

Differentiate again (and again)

The derivative f'(x) of a function is itself a function — so we can differentiate it too. Doing that gives the second derivative, written f''(x). Keep going and you get the third, f'''(x), and so on, as far as you like:

f \;\xrightarrow{\;\frac{d}{dx}\;}\; f' \;\xrightarrow{\;\frac{d}{dx}\;}\; f'' \;\xrightarrow{\;\frac{d}{dx}\;}\; f''' \;\cdots

Each one measures the rate of change of the one before it: f' is how fast f changes, f'' is how fast f' changes, f''' is how fast f'' changes. There is nothing new to learn: you already know how to differentiate a polynomial — you just do it again. The skill is entirely one of bookkeeping, so most of this page is practice in keeping the books straight.

Three notations are in circulation, and you will meet all of them:

f''(x) \;=\; \frac{d^2y}{dx^2} \;=\; \ddot{x} \quad\text{(Lagrange, Leibniz, Newton)}

The Leibniz form looks alarming but just records "apply \tfrac{d}{dx} twice": \tfrac{d}{dx}\!\left(\tfrac{dy}{dx}\right) = \tfrac{d^2y}{dx^2}. Newton's dots (\dot{x}, \ddot{x}) are the physicists' shorthand when the variable is time. Past the third derivative the prime marks get unreadable, so we switch to f^{(n)}(x) — the nth derivative, with the n in parentheses so it can't be mistaken for a power.

Watch the powers march down

Take f(x) = x^4. Apply the power rule over and over and the powers march downward, each step pulling its exponent out front as a factor:

f(x) = x^4,\quad f'(x) = 4x^3,\quad f''(x) = 12x^2,\quad f'''(x) = 24x,\quad f^{(4)}(x) = 24

After four steps the x is gone and we hit the constant 24 = 4\cdot3\cdot2\cdot1 = 4!; one more derivative and everything is 0 — and stays 0 forever. Every polynomial dies this way: each differentiation lowers the degree by exactly one, so a degree-4 polynomial survives four differentiations, becomes a constant, and vanishes on the fifth. That degree-countdown is the single most useful fact on this page — it tells you, before you compute anything, what shape each answer must have.

Let f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 be a polynomial of degree n. Then:

Worked example 1: a full polynomial, all the way down

Let's take a degree-4 polynomial with all the trimmings and differentiate it to extinction. Watch two things as we go: each term obeys the power rule independently, and the degree drops by one at every step.

f(x) = 2x^4 - 3x^3 + 5x - 7

First derivative. Power rule term by term — the constant -7 dies immediately:

f'(x) = 8x^3 - 9x^2 + 5 \qquad \text{(degree } 4 \to 3\text{)}

Second derivative. Differentiate f'. Now the +5 is the constant, and it dies:

f''(x) = 24x^2 - 18x \qquad \text{(degree } 3 \to 2\text{)}

Third derivative.

f'''(x) = 48x - 18 \qquad \text{(degree } 2 \to 1\text{)}

Fourth and beyond.

f^{(4)}(x) = 48, \qquad f^{(5)}(x) = 0, \qquad f^{(6)}(x) = 0, \;\ldots

Sanity checks: the degrees ran 4, 3, 2, 1, 0 — no skips. And the final constant is 48 = 4! \cdot 2, exactly the theorem's n!\,a_n with n = 4 and leading coefficient a_n = 2. If your fourth derivative of a quartic still has an x in it, a step went wrong.

Worked example 2: a higher derivative at a point

Exam questions usually want a number, not a formula: "find f''(3)". The order of operations matters — differentiate first, substitute last. Take

f(x) = x^3 - 4x^2

Step 1 — build the derivatives as formulas.

f'(x) = 3x^2 - 8x, \qquad f''(x) = 6x - 8, \qquad f'''(x) = 6

Step 2 — now substitute x = 3.

f''(3) = 6(3) - 8 = 10, \qquad f'''(3) = 6

Notice f''' came out as the constant 6 — so f'''(3) = f'''(100) = f'''(-\pi) = 6. The third derivative of any cubic ax^3 + \cdots is the constant 6a, everywhere. (Substituting first — computing f(3) = -9 and then "differentiating" the number — gives 0, which is nonsense: a number doesn't change, but the function around x=3 certainly does.)

See all three stacked together

A higher derivative isn't just algebra — each one is a curve you can read. Below is f(x) = x^3 - 3x (bold) with its first derivative f'(x) = 3x^2 - 3 and second derivative f''(x) = 6x on the same axes. Slide the tangent line along the bold curve and check the chain of readings with your own eyes:

Try x = 0: the tangent slopes down (f'(0) = -3) yet f''(0) = 0 — the curve is neither bending up nor down at that instant. The sign of f'' tells you which way a curve bends, and that observation grows into a whole tool of its own in concavity.

Worked example 3: the motion chain s \to v \to a

If s(t) is the position of a moving object at time t, then its derivative is velocity and the derivative of velocity is acceleration:

v(t) = s'(t), \qquad a(t) = v'(t) = s''(t)

So acceleration is the second derivative of position — the mathematical name for that push you feel in the car seat. Let's run the whole chain with numbers. A particle moves along a line with position (metres, seconds)

s(t) = t^3 - 6t^2 + 9t

Differentiate twice:

v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3), \qquad a(t) = s''(t) = 6t - 12

Now read the story off the numbers:

One cubic, two differentiations, and you can narrate the particle's entire journey — forward, stall, reverse, stall, forward — without plotting a single point. That is what higher derivatives are for.

Watch the chain in motion

Press play to watch a ball thrown straight up. Its height follows s(t) = -5t^2 + 20t, so the chain gives v(t) = s'(t) = -20t + 20 and a(t) = s''(t) = -10. Watch how the three behave completely differently: the position traces a parabola, the velocity drains steadily from positive through zero (at the very top) to negative — and the acceleration never changes at all. Gravity pulls down by the same amount the whole flight, even at the top where the ball hangs still for an instant. A constant second derivative with a changing first derivative: the quadratic's signature.

This is why a quadratic models free fall: "constant acceleration" means s'' = \text{const}, and the only functions whose second derivative is constant are the quadratics.

The second-derivative notation is a minefield of near-misses. These three are not the same function — check them on f(x) = x^3:

The same trap hides inside Leibniz notation. In \dfrac{d^2y}{dx^2} the little 2s are counters, not powers — they record "the operator \tfrac{d}{dx} applied twice". So

\frac{d^2y}{dx^2} \;\neq\; \left(\frac{dy}{dx}\right)^{\!2}

For y = x^3: the left side is 6x, the right side is 9x^4. Notice even the placement encodes this: the superscript sits on the d upstairs (d^2y, "two d's applied to y") but on the whole dx downstairs (dx^2 means (dx)^2). A quick self-test before any exam: if your "second derivative" of a cubic has degree 4, you squared when you should have differentiated.

The chain doesn't stop at acceleration. The derivatives of position have official names as far as the sixth:

s \to \underbrace{v}_{\text{velocity}} \to \underbrace{a}_{\text{acceleration}} \to \underbrace{j}_{\text{jerk}} \to \underbrace{s^{(4)}}_{\text{snap}} \to \underbrace{s^{(5)}}_{\text{crackle}} \to \underbrace{s^{(6)}}_{\text{pop}}

Jerk — the third derivative — is the one your body objects to. Constant acceleration feels like a steady, predictable push (astronauts tolerate several g of it for minutes). What makes you stumble on a train or feel queasy in a lift is the push changing: high jerk. That's why lift engineers and roller-coaster designers publish jerk limits, not just acceleration limits. Early looping coasters used circular loops, and the acceleration jumped so abruptly at the loop's entry — enormous jerk — that they injured riders' necks. Modern loops are teardrop-shaped clothoid curves, engineered so the acceleration ramps up smoothly: the designers are deliberately taming a third derivative. Snap, crackle and pop (named, wonderfully, after the Rice Krispies elves) matter in precision engineering — camera gimbals and high-speed rail track geometry are designed with snap in mind — though by then the effects are subtle indeed.

Watch it on Khan Academy

A second voice on the same idea — how the second derivative is computed and what it says about a graph:

See it explained