1The idea in one paragraph

Differentiate a function and you get its gradient function f′, which says how steep the graph is at each x. Differentiate again and you get the second derivative f″, which says how fast that gradient is changing. When f″ is positive the gradient is increasing, and the graph bends upwards like a cup: it is concave up. When f″ is negative the gradient is decreasing, and the graph bends downwards like a cap: it is concave down. Where the bending changes over, the graph has a point of inflexion. So one function carries three graphs. The graph of f shows the values. The graph of f′ shows the slopes. The graph of f″ shows the bending. The zeros of one graph line up with the turning points of the graph before it, and reading those alignments is most of this page.

2Differentiating twice

The second derivative is the derivative of the derivative. There are two notations, and the guide expects you to use both.

• Leibniz notation: if y is a function of x, the second derivative is written d²y/dx², read "d two y by d x squared". It means d/dx (dy/dx): the operation d/dx done twice, which is why the 2 sits on the d at the top and on the x at the bottom.
• Function notation: if the function is f, the second derivative is f″(x), read "f double dash of x".

Neither notation means the first derivative squared. (dy/dx)² is a different thing entirely.

Start with a polynomial.

A composite needs the chain rule each time. Every differentiation of sin 3x brings out another factor of 3.

A product usually stays a product after one differentiation, so the product rule is needed again. Take f(x) = x e^2x.

Factorising f′ before differentiating again is a habit worth building. It keeps the algebra short, and a factorised f″ is exactly what you need in section 4 to find its sign. As a check, f″(0) = 4(1)(1) = 4.

A quotient needs the quotient rule twice, and that is where Paper 1 sets its "show that" questions: Q2 in Try it is one.

On Paper 2 a GDC can evaluate f″ at a point numerically. That gives a number, not a formula: it answers "find f″(2)", never "find f″(x)".

3What the sign of f″ means: concavity

f″ is the rate of change of f′. So:

f″(x) > 0: the gradient is increasing, and the graph is concave up. f″(x) < 0: the gradient is decreasing, and the graph is concave down.

Figure 1 shows why the words fit. On the left, the tangent gradients go −2, 0, 2 from left to right: climbing, so the curve bends up into a cup. On the right they go 2, 0, −2: falling, so the curve bends into a cap. Notice too that a concave-up curve lies above its tangents and a concave-down curve lies below them.

Concavity and direction are separate questions: f′ says up or down, f″ says which way the graph bends, and all four combinations occur, as Figure 2 shows. A graph can rise and be concave down (it is climbing but flattening out, like a ball rising towards the top of its flight) or fall and be concave up (it is dropping but levelling off, like a cup of tea cooling towards room temperature).

In context, the second derivative describes the growth itself. If P(t) is the number of users of a new app, P′(t) > 0 says the number is growing; P″(t) > 0 says growth is speeding up, and P″(t) < 0 says the app is still gaining users but more slowly each week. When a question says "interpret P″(t) < 0", that sentence, in context, is the answer.

4Finding where a graph is concave up or concave down

This is the method of 5.2 for increasing and decreasing, applied one derivative later.

1. Find f″(x) and factorise it fully.
2. Solve f″(x) = 0. These values split the number line into intervals.
3. Find the sign of f″ on each interval, using the factors or a test value.
4. Write the intervals with f″ > 0 (concave up) and f″ < 0 (concave down).

Example 1, a cubic. f(x) = x³ − 6x² + 9x + 1.

A cubic's second derivative is linear, so every cubic changes concavity exactly once. Section 5 draws all three graphs of this one.

Example 2, a quartic. f(x) = x⁴ − 6x².

So the graph is concave up for x < −1 and for x > 1, and concave down for −1 < x < 1. Figure 3 colours the curve by concavity. The bend changes at (−1, −5) and (1, −5), and these are the points of inflexion that 5.8 classifies.

Example 3, a factor that never changes sign. f(x) = x e^−x.

Because e^−x > 0 for every x, the sign of f″ is the sign of x − 2 alone. The graph is concave down for x < 2 and concave up for x > 2. Say the reason in your answer ("since e^−x > 0"): on a "justify" question it is the R1 mark.

Use strict inequalities for concavity intervals. At the point where f″ = 0 itself the graph is neither concave up nor concave down; it is switching.

5One function, three graphs

Now put the three graphs on top of each other. Figure 4 stacks f(x) = x³ − 6x² + 9x + 1, its derivative f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3), and f″(x) = 6x − 12, with the x-axes lined up.

Read down each dashed line.

• At x = 1 f has a local maximum, (1, 5), so f′(1) = 0 and f′ crosses the axis from positive to negative, because f goes from rising to falling.
• At x = 3 f has a local minimum, (3, 1), and f′ crosses from negative to positive.
• At x = 2 f is at its steepest downhill: the gradient f′(2) = −3 is the lowest value f′ takes. So the graph of f′ has its minimum here, the graph of f″ crosses zero, and the graph of f changes from concave down to concave up at (2, 3). That is the point of inflexion.

Every one of those observations is a line of this table.

The pattern is a ladder: each graph's zeros sit under the turning points of the graph above it. The ladder fixes the shape of f but not its height, since f and f + 5 have the same derivative: the constant of integration from 5.5, seen as a picture.

6Sketching f′ from f, and reading f from f′

Exams test the table from both ends.

From the graph of f to the graph of f′. Work through it feature by feature.

1. Mark every stationary point of f. Each one is a zero of f′.
2. Between those zeros, decide whether f is going up (f′ above the axis) or down (f′ below).
3. Mark the points of inflexion of f. Each one is a turning point of f′, at the height of the gradient there.
4. At the ends: if f gets ever steeper, f′ heads off to ±∞; if f levels off, f′ heads towards 0.

Figure 5 does this for f(x) = x⁴ − 4x³, whose derivative is f′(x) = 4x³ − 12x² = 4x²(x − 3). At x = 3 the graph of f turns (a minimum at (3, −27)), so f′ crosses the axis from negative to positive. At x = 0 the graph of f flattens but carries on falling: f′ is negative just before 0, zero at 0, and negative just after. So the graph of f′ touches the x-axis at 0 without crossing it, as the squared factor x² predicts. The steepest downhill point of f, at x = 2 where f′(2) = −16, becomes the minimum of f′.

From the graph of f′ back to f. This is the more common exam form, and the trap is to read the graph as if it were f. Write "this is f′" next to it first, then use the table from right to left.

Figure 6 shows the graph of y = f′(x) for a function f. You do not need its formula, but it is f′(x) = (x + 1)(x − 2)².

• At A, x = −1, f′ crosses the axis from negative to positive. So f changes from decreasing to increasing: f has a local minimum at x = −1.
• At C, x = 2, f′ touches the axis and stays positive on both sides. So f is stationary at x = 2 but keeps increasing through it. This is not a maximum or a minimum; it is a stationary point of inflexion, which 5.8 names properly.
• At B, x = 0, f′ has a local maximum. The gradient of f stops increasing and starts decreasing, so f changes from concave up to concave down: a point of inflexion of f, where the gradient is f′(0) = 4.
• Concavity: f′ is increasing for x < 0 and for x > 2, so f is concave up there, and decreasing for 0 < x < 2, so f is concave down there. C is both a zero and a turning point of f′, which is why the stationary point at x = 2 is also a point of inflexion.

7Letting the GDC find the zeros of f″

Some second derivatives cannot be solved by hand, and on Paper 2 that is no obstacle. Take f(x) = eˣ − x³.

Graph y = eˣ − 6x on the GDC and use the zero (root) tool twice. Figure 7 shows the result: x ≈ 0.204 and x ≈ 2.83. Reading the sign of f″ off the graph between those zeros:

• f″ > 0 for x < 0.204 and for x > 2.83, so f is concave up there;
• f″ < 0 for 0.204 < x < 2.83, so f is concave down there.

Write down what you graphed: "graph of y = eˣ − 6x, zeros at x = 0.204 and x = 2.83" earns the method mark that a bare interval does not. And if you cannot find f″ by hand at all, graph the numerical derivative of f: its turning points are the points of inflexion of f.

The guide links this topic to simple harmonic motion in physics. If a mass on a spring has displacement x = 3 cos 2t, then d²x/dt² = −12 cos 2t = −4x: the acceleration is proportional to the displacement and opposite in sign, which is what defines simple harmonic motion. In 5.9 the second derivative of displacement is acceleration.