Second derivative

What it is

The second derivative f″ comes from differentiating twice: first f to f′, then f′ to f″. f′(x) tells how steeply the graph slopes at x, and f″(x) tells how that slope is changing there. The second derivative of y = f(x) is written f″(x), y″ or d²y/dx².

Example: f(x) = x³ − 3x has f′(x) = 3x² − 3 and f″(x) = 6x.

The sign of f″ tells which way the graph bends:

• Where f″ > 0 the slope increases, and the graph bends upwards like a bowl. It is called concave up, or convex.
• Where f″ < 0 the slope decreases, and the graph bends downwards like a roof. It is called concave down, or concave.
• If f″ = 0 on a whole interval the graph does not bend there at all. f(x) = 3x + 1 has f′ = 3 and f″ = 0, and its graph is a straight line.

A point where f″ changes sign, so that the graph goes from bending one way to bending the other, is called an inflection point. x³ − 3x has one at (0, 0), where f″ goes from negative to positive.

Some second derivatives worth knowing:

Not an inflection point: f(x) = x⁴ has f″(x) = 12x². It is 0 at x = 0 but positive on both sides, so the graph bends upwards through the origin too.

Why it matters

• Maxima and minima. At a stationary point, where f′(a) = 0, the sign of f″(a) decides whether the point is a minimum (f″(a) > 0) or a maximum (f″(a) < 0); see the second derivative test. When f″(a) = 0 the test gives no answer, and the first derivative test takes over. Example: g(x) = x² − 4x + 1 has g′(x) = 2x − 4, which is 0 at x = 2. There g″(x) = 2 > 0, so x = 2 is a minimum with g(2) = −3. Finding the stationary points is covered on the stationary points page.
• Optimisation. A rectangle with perimeter 40 m has sides x and 20 − x and area A(x) = 20x − x². A′(x) = 20 − 2x is 0 at x = 10, and A″(x) = −2 < 0 shows it is a maximum: the largest area is 100 m².
• Acceleration. If s(t) is position, s′(t) is velocity and s″(t) is acceleration; see acceleration as a derivative. For s(t) = t³ − 6t² + 9t, in metres and seconds, s′(t) = 3t² − 12t + 9 and s″(t) = 6t − 12. At t = 1 the velocity is 0 and the acceleration −6 m/s², and the position has a local maximum, 4 m. At t = 3 the velocity is 0 and the acceleration 6 m/s², and the position has a local minimum, 0 m.
• Approximations. The Taylor polynomial of degree 2 uses f″(a): f(x) ≈ f(a) + f′(a)(x − a) + f″(a)/2 · (x − a)².

The derivative calculator shows both f′ and f″, and the graphing calculator shows how the graph bends.

Common misconceptions

• “f″(x) is the same as (f′(x))².” f″ means differentiating twice. For f(x) = x³, f″(x) = 6x, but (f′(x))² = (3x²)² = 9x⁴. In the same way, d²y/dx² is not (dy/dx)².
• “A positive second derivative means the function is increasing.” f″ tells how the graph bends, not which way it goes. f(x) = x² has f″(x) = 2 > 0 everywhere but decreases for x < 0. Conversely, ln x increases for every x > 0 although f″(x) = −1/x² is negative.
• “If f″(a) = 0, a is an inflection point.” f″ also has to change sign. f(x) = x⁴ has f″(0) = 0 but no inflection point at 0.
• “The second derivative of a straight line is its slope.” The slope is f′. For f(x) = 3x + 1, f′ = 3 and f″ = 0, because a straight line does not bend.

Common questions