A third derivative calculator differentiates a function three successive times, producing f‴(x). It is the derivative most students meet last: less famous than the slope and concavity measures below it, but the one that describes how curvature itself is changing — and, in motion problems, the quantity engineers call jerk.
Quick Answer
Differentiate three times in a row: f → f′ → f″ → f‴. For f(x) = x⁵ − 4x³ + 2x: f′(x) = 5x⁴ − 12x² + 2, f″(x) = 20x³ − 24x, and f‴(x) = 60x² − 24. At x = 1 the third derivative equals 36; at x = 0 it equals −24.
What a Third Layer of Change Means
Read the ladder physically. If f is position, f′ is velocity (rate of change of position), f″ is acceleration (rate of change of velocity), and f‴ is jerk — the rate of change of acceleration. Jerk is the difference between a lift that starts smoothly and one that yanks: passengers feel jerk, not acceleration alone, which is why ride designers and robotics engineers genuinely compute this quantity. In pure curve terms, f‴ measures how the concavity is evolving; where f″ has its own peaks and troughs, f‴ is zero, and f‴ is also the referee in the borderline inflection case — if f″(a) = 0 and f‴(a) ≠ 0, the point is certainly an inflection point.
Notation causes real confusion here, so meet the main forms once: f‴(x) (three primes), d³y/dx³ (Leibniz form — the 3s are orders, not powers), and f⁽³⁾(x) with the order in brackets. None of them means f cubed or the third power of anything. Polynomial behaviour is predictable: each differentiation lowers the degree by one, so any polynomial of degree less than 3 — lines and quadratics — has a third derivative of exactly zero, and the third derivative of a cubic is a constant.
For the curious, the ladder does not stop at three. The fourth, fifth, and sixth derivatives of position carry the informal names snap, crackle, and pop — coined half-jokingly after the cereal mascots, but used seriously enough in motion-control work, where limiting snap in a robot arm's path is a genuine design constraint. You will rarely need them, yet they underline the point of this whole family: every derivative answers "how fast is the previous answer changing?", and the ladder ends only when the answer becomes zero or the question stops being useful.
How to Use the Third Derivative Calculator
- Enter the function, expanding or simplifying it first if that is how you are working by hand.
- Read the third derivative returned.
- If you need the value at a point, substitute the x-value into f‴.
- To locate where jerk changes sign or where f″ peaks, solve f‴(x) = 0.
- Check your own work one rung at a time — verify f′ before f″, and f″ before f‴.
- For trigonometric or exponential functions, watch for the derivative cycles (each round returns to a familiar form).
Worked Example: Three Rounds on a Quintic
Start with f(x) = x⁵ − 4x³ + 2x. Round one, the power rule term by term: f′(x) = 5x⁴ − 12x² + 2. Round two: f″(x) = 20x³ − 24x. Round three: f‴(x) = 60x² − 24. Now evaluate: at x = 1, f‴ = 60 − 24 = 36; at x = 0.5, f‴ = 60(0.25) − 24 = 15 − 24 = −9. The third derivative vanishes where 60x² = 24, i.e. x² = 0.4, so x ≈ ±0.63 — the points where f″ stops increasing and starts decreasing, or the reverse. Other standard third derivatives show the patterns to expect:
- f(x) = sin x → f‴(x) = −cos x (the four-step cycle: sin, cos, −sin, −cos)
- f(x) = e^(2x) → f‴(x) = 8e^(2x)
- f(x) = x⁴ → f‴(x) = 24x
- f(x) = x³ → f‴(x) = 6, a constant, as promised for cubics
The physics ladder deserves one concrete set of numbers. Suppose a cart's position along a track is s(t) = t³ − 3t² + 2t metres. Its velocity is s′(t) = 3t² − 6t + 2 and its acceleration is s″(t) = 6t − 6, both changing with time — but the jerk, s‴(t) = 6, is a constant 6 m/s³: the acceleration itself ramps up at a perfectly steady rate. At t = 2 seconds the cart sits at s = 8 − 12 + 4 = 0 m (back at the sensor), moving at s′ = 12 − 12 + 2 = 2 m/s, with acceleration s″ = 6 m/s² and jerk 6 m/s³. A jerk that stays constant is the smooth, predictable case; it is a jerk that spikes — when a driver stamps on a brake rather than pressing it — that passengers feel in their stomachs, and that vehicle engineers design whole control systems to keep small.
Errors That Compound Over Three Rounds
The defining hazard of this topic is inheritance: a coefficient dropped in f′ is wrong in f″ and wrong again in f‴, and the final answer can look plausible while being wrong in every term — which is why verification rung by rung is the whole game. The second error is conceptual: reaching for f‴ to find inflection points directly. Inflection candidates come from f″(x) = 0; the third derivative only tests those candidates, and using it alone finds the wrong points (in the example, f‴ = 0 at ±0.63, while the inflection point of f sits where f″ = 20x³ − 24x = 0 — different equation, different answers). Third, prime-counting slips in notation: f⁽³⁾ written as f³ changes the meaning entirely, and in Leibniz form d³y/dx³ the exponent belongs to the operator and to dx as a unit, never to y. Fourth, with products and compositions the lower rules still apply at every round — the third derivative of sin(2x) needs the chain rule three times, and missing the factor of 2 once corrupts the rest.
Stepping Back Down the Ladder
Most questions that reach f‴ start life one rung lower, so keep the Second Derivative Calculator open for concavity and the second derivative test, and the Derivative Calculator for the foundational first step — comparing a hand-worked chain against all three tools is a thorough way to find which rung a mistake crept in on. The complete differentiation family is gathered in the Math Calculators category.