Second Derivative Calculator

What the second derivative measures

The first derivative gives the slope of a curve at a point. The second derivative gives the rate at which that slope changes. Take a car: its position in miles is f(t), its speed is f'(t) in miles per hour, and its acceleration is f''(t) in miles per hour per hour. MIT OpenCourseWare's session on higher derivatives (18.01SC, Fall 2010) introduces f'' exactly this way, as the derivative of the derivative, with the notation d2y/dx2 for the result.

On a graph the sign of f'' describes the bend. Where f'' is positive the slope is increasing, so the curve opens upward like a cup. Where f'' is negative the slope is decreasing and the curve opens downward. A straight line has f'' = 0 everywhere, which is why a linear function has no bend to report.

Units carry through the second differentiation, and that is where most confusion starts. A height in feet against time in seconds gives f' in feet per second and f'' in feet per second squared. A revenue model in dollars against a price in dollars gives f'' in dollars per dollar squared, a number that answers a different question than the first derivative.

Differentiate twice, rule by rule

Nothing new is needed for the second pass. Apply the same rules to f' that produced f' from f: the power rule, the constant multiple rule, the sum and difference rules, the product rule, the quotient rule and the chain rule. A constant term disappears in the first pass, and a linear term disappears in the second.

Take f(x) = 5x^4 - 3x^2 + 7x - 2. The first derivative is 20x^3 - 6x + 7 and the second is 60x^2 - 6. The -2 is gone after one step, and the 7x is gone after two. OpenStax Calculus Volume 1, section 3.3 lists the same rules in the order most courses teach them, with the constant multiple and sum rules stated before the product and quotient rules.

Each rule leaves a mark on the second derivative. The power rule lowers the exponent twice, so x^5 becomes 5x^4 and then 20x^3. The product rule adds a term at each step, so a product of two functions usually has three terms in its second derivative. The chain rule brings the inner derivative down twice, which is why sin(2x) has f'' = -4sin(2x) and not -sin(2x).

Worked example: x^3 - 2x^2 + 3x at x = 2

Work term by term. The cube gives 3x^2, the -2x^2 gives -4x, and 3x gives 3, so f'(x) = 3x^2 - 4x + 3. Differentiate again: 3x^2 gives 6x, -4x gives -4, and the 3 drops out, so f''(x) = 6x - 4.

At x = 2 the three values are f(2) = 8 - 8 + 6 = 6, f'(2) = 12 - 8 + 3 = 7 and f''(2) = 12 - 4 = 8. A positive second derivative means the curve is concave up there, and the slope is climbing by about 8 units for every unit of x.

Test that reading with two nearby points. f'(1.9) = 3(3.61) - 7.6 + 3 = 6.23, and f'(2.1) = 3(4.41) - 8.4 + 3 = 7.83. The slope rose 1.6 over a run of 0.2, an average of 8 per unit, which matches f''(2) = 8. That is the whole idea of a second derivative in one check: it is the slope of the slope, measured between two nearby slopes.

The second derivative test for maxima and minima

Set f' = 0 to find the flat spots, then read f'' at each one. If f'' is positive the curve is concave up, so the flat spot is a local minimum. If f'' is negative the curve is concave down and the flat spot is a local maximum. If f'' is zero the test says nothing, and you have to check the sign of f' on either side of the point.

Example: f(x) = x^3 - 3x. Then f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1), so the critical points are x = 1 and x = -1. The second derivative is f''(x) = 6x, which is 6 at x = 1 and -6 at x = -1. So f(1) = -2 is a local minimum and f(-1) = 2 is a local maximum.

The same test shows up in business models. If profit is P(q) = -2q^2 + 40q - 150, then P'(q) = -4q + 40 is zero at q = 10 and P''(q) = -4 everywhere, so 10 units is a maximum rather than a minimum. A positive second derivative on a cost curve tells the opposite story: cost per unit is rising faster and faster as output grows.

Concavity, inflection points and the sign chart

Concavity is a property of an interval, not of a single point. For f(x) = x^3 the second derivative is f''(x) = 6x, negative for every x below zero and positive for every x above zero. The curve is concave down on the left, concave up on the right, and the point (0, 0) where the sign flips is an inflection point.

A zero second derivative alone does not prove an inflection. For f(x) = x^4, f''(x) = 12x^2, which is zero at x = 0 but positive on both sides, so the curve stays concave up and there is no inflection. The test needs a sign change across the point, not just a zero.

Build a sign chart by finding where f'' is zero or undefined, then testing one value in each interval. For f(x) = x^3 - 3x, f''(x) = 6x is zero only at 0, so the chart has two intervals. LibreTexts Calculus, section 4.5 states the concavity test in the same form and works through sign charts for several functions.

Acceleration: the second derivative you can feel

A ball thrown upward from 5 feet with a speed of 64 feet per second has height h(t) = -16t^2 + 64t + 5, where the -16 is half of the 32 feet per second squared acceleration due to gravity. Then h'(t) = -32t + 64 and h''(t) = -32.

The constant second derivative says gravity does not change: the speed drops by 32 feet per second every second, whether the ball is rising or falling. At t = 0 the ball is at 5 feet with a speed of 64 ft/s. At t = 1 it is at 53 feet moving up at 32 ft/s. At t = 2 it is at 69 feet with a speed of zero, the top of the flight. The second derivative was -32 the whole time, and that is what makes the path a parabola rather than some other curve.

Solving h(t) = 0 gives t = (64 + sqrt(4,096 + 320)) / 32 = 4.08 seconds, so the ball lands about 4.08 seconds after the throw. Its speed at that moment is h'(4.08) = -32(4.08) + 64 = -66.6 ft/s, the same 64 ft/s it left with plus a little, because it started 5 feet above the ground.

How this calculator reads your function

Type x as the variable and use ^ for powers, so x^3 means x cubed. A number written next to x means multiplication, so 3x and 2x^2 both work. Function names take parentheses: sin(x), cos(2x), exp(x), ln(x), log(x) for base 10, sqrt(x) and abs(x). The constants pi and e are built in.

For polynomials and rational functions the calculator keeps exact coefficients and cancels common factors, so 1/x returns f''(x) = 2/x^3 instead of a stack of unsimplified terms. Functions such as sin, exp and ln are differentiated with the standard rules and then evaluated at the point you entered.

Some inputs have no value at the chosen point. ln(x) at x = 0, 1/x at x = 0 and sqrt(x) below zero return a message instead of a number, because the function itself is undefined there. A second derivative can also fail where the first derivative has a corner, as with abs(x) at zero, and the panel says so rather than printing a wrong value.

Two display habits are worth knowing. The derivative formulas are shown before the numbers, so you can copy the algebra into homework, and the point values are rounded to ten significant digits. Rounding only affects the printed number, never the differentiation itself.

Mistakes that change the answer

• Dropping the inner derivative. sin(2x) has f'' = -4sin(2x), not -sin(2x), because the chain rule applies on both passes.
• Losing a sign on cosine. cos(x) gives f' = -sin(x) and f'' = -cos(x); the minus survives both steps.
• Confusing f'' with the square of f'. For f(x) = x^2, f'' = 2 while (f')^2 = 4x^2. They are different functions with different units.
• Treating f'' = 0 as an inflection point. x^4 has f'' = 0 at the origin and no inflection, because the sign never changes.
• Reading a point value as an interval property. f''(2) = 8 says the curve bends up at x = 2, not that it bends up everywhere.
• Mixing units halfway through. A height in feet and time in seconds gives feet per second squared. Converting the height to meters without converting the time changes the second derivative by the length conversion factor.