A second derivative calculator differentiates a function twice, producing f″(x) — the derivative of the derivative. Where the first derivative measures slope, the second measures how that slope itself is changing: whether a curve is bending upward or downward, where it flips from one to the other, and whether a flat point is a peak, a trough, or neither.
Quick Answer
Find f′(x), then differentiate once more. For f(x) = x³ − 6x² + 9x + 1: f′(x) = 3x² − 12x + 9 and f″(x) = 6x − 12. The sign of f″ gives the concavity — negative for x < 2 (concave down), positive for x > 2 (concave up) — and f″ = 0 at x = 2 marks the inflection point.
Concavity, in Plain Terms
A curve is concave up where it holds water — the slope is increasing, so the graph bends upward like a bowl. It is concave down where it sheds water, the slope decreasing, like a cap. The second derivative is the instrument that detects which regime you are in, because an increasing first derivative means a positive second derivative. An inflection point is where the bending switches sides; f″ must be zero or undefined there, but — and this matters — not every zero of f″ is an inflection point. The bending must actually change sign.
The second famous job of f″ is the second derivative test for classifying stationary points. Where f′(a) = 0: if f″(a) > 0 the curve is bowl-shaped at that spot, so a is a local minimum; if f″(a) < 0 it is cap-shaped, a local maximum; if f″(a) = 0 the test says nothing at all, and the point could be either, or an inflection point instead — fall back on examining the sign of f′ on both sides. In physics the same quantity is acceleration when the original function is position, which is why f″ keeps appearing far outside pure mathematics.
How to Use the Second Derivative Calculator
- Enter the function in its simplest form — expand brackets first if your hand working does.
- Read the second derivative returned.
- To test concavity on an interval, substitute any point from that interval into f″ and note the sign.
- To hunt for inflection points, solve f″(x) = 0 and then check that f″ changes sign across each candidate.
- To classify a stationary point, first confirm f′ equals zero there, then read the sign of f″ at that same x.
- Compare against your own two rounds of differentiation line by line if the answers disagree.
Worked Example: Peaks, Troughs, and the Turn Between Them
Take f(x) = x³ − 6x² + 9x + 1. The first derivative f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3) is zero at x = 1 and x = 3. Apply the second derivative test with f″(x) = 6x − 12: at x = 1, f″ = −6, negative, so the point (1, f(1)) = (1, 5) is a local maximum; at x = 3, f″ = +6, positive, so (3, f(3)) = (3, 1) is a local minimum. Halfway between them, f″(2) = 0 and the sign of f″ flips from negative to positive, so (2, f(2)) = (2, 3) is a genuine inflection point — the curve stops capping and starts bowling exactly where it should, midway between the peak and the trough. Companion facts worth knowing:
- f(x) = x² has f″(x) = 2 everywhere: always concave up
- f(x) = ln x has f″(x) = −1/x²: concave down across its whole domain
- f(x) = x⁴ has f″(0) = 0 yet no inflection at 0, because f″(x) = 12x² never changes sign
- A straight line has f″(x) = 0 everywhere — no bending at all
Read the worked example as a motion problem and the test becomes intuitive. Treat f as the position of a particle, so f′ is its velocity and f″ its acceleration. At x = 1 the velocity is zero and the acceleration is −6: the particle is momentarily stopped and being pulled backwards, which is precisely what a local maximum of position looks like in motion — the furthest point reached before returning. At x = 3 the velocity is zero again but the acceleration is +6, pushing the particle forward: the local minimum, the nearest point before it sets off again. At x = 2 the acceleration passes through zero and changes sign — the inflection point — which in motion is the instant the pull switches from backward to forward. Students who anchor the signs to this picture rarely confuse maxima with minima again, because the physics refuses to let them.
Misreadings of the Second Derivative
The costliest misconception is treating every solution of f″(x) = 0 as an inflection point; the x⁴ example above shows a zero with no change of concavity, so always run the sign check. The second is applying the second derivative test to a point that is not stationary — the test is meaningless unless f′(a) = 0 first. Third, when f″(a) = 0 at a stationary point, students often declare it an inflection point automatically; in fact the test has simply failed, and a first-derivative sign chart must decide (x³ at 0 is an inflection point, x⁴ at 0 is a minimum — both have f′ = f″ = 0 there). Fourth, arithmetic: two rounds of differentiation double the chances of a dropped coefficient, so verify f′ carefully before differentiating it — an error in round one is inherited, silently, by round two.
One Derivative Further
If a problem pushes past f″, the Third Derivative Calculator continues the chain — and can even rescue the inflection hunt, since f″(a) = 0 together with f‴(a) ≠ 0 guarantees an inflection point. For the underlying first step, the Derivative Calculator handles f′ itself, and the Extreme Points Calculator automates the peak-and-trough search that the second derivative test performs by hand. All of them are shelved together in the Math Calculators category.