1The idea in one paragraph

On a curve the gradient keeps changing, so you cannot find it with two far-apart points. But you can take two points very close together. The gradient of the line joining them, a chord, is nearly the gradient of the curve there, and the closer the points, the better the approximation. Push the second point towards the first and the chord gradients settle on one number. That settling-down value is a limit, and the number it gives is the derivative: the gradient of the curve at that point, and equally the rate at which one quantity is changing with respect to another at that instant. Do this at every point and you get a whole new function, the gradient function f′(x).

2What a limit is

A limit is the value a function's output gets closer and closer to as its input gets closer and closer to some number, whether or not the function actually reaches that value, and whether or not the function is even defined there.

We write it like this:

lim_x→a f(x) = L · "as x tends to a, f(x) tends to L". It says what happens near a, and nothing about what happens at a.

That last sentence is the whole point. Take

f(x) = (x² − 4) ÷ (x − 2).

Try to substitute x = 2 and you get 0 ÷ 0, which has no value. So f(2) does not exist. But ask what happens near 2 and there is a perfectly clear answer.

From the left and from the right, the outputs close in on 4. So lim_x→2 f(x) = 4, even though f(2) does not exist.

Figure 1 shows why. For every x except 2, the top factorises as (x − 2)(x + 2), the (x − 2) cancels, and f(x) = x + 2. The graph is the line y = x + 2 with a single point missing at (2, 4): a hole. The limit is the height of the hole.

Two facts from this example carry through the whole topic.

Approach from both sides. A limit exists only if the values from the left and the values from the right head for the same number. In the table, check both halves.

The limit need not equal the value. Here f(2) does not exist at all. For the smooth functions you will differentiate, the limit and the value usually do agree, but the definition of a limit never uses the value at the point, and derivatives depend on exactly that.

3Estimating a limit from a table or a graph

The guide asks you to estimate limits, not to calculate them by formal methods; formal methods are not required at standard level. Estimating means choosing inputs closer and closer to the target, reading the outputs, and saying what they approach.

From a table. Take y = sin x ÷ x, with x in radians. You cannot cancel anything this time, and x = 0 gives 0 ÷ 0 again. So build a table, which on Paper 2 is a job for the GDC's table function.

The values approach 1 from both sides, so lim_x→0 (sin x ÷ x) = 1. Figure 2 draws the function, with the hollow circle at the point it approaches but never reaches.

Three habits make an estimate reliable:

• Use inputs that get much closer each time, such as 0.1, 0.01, 0.001, not 0.1, 0.09, 0.08.
• Check both sides of the target.
• State the limit as the value the outputs approach, not as the last number in your table. The last entry above is 0.99998, but the limit is 1.

From a graph. Follow the curve towards the x-value from each side with your finger and read the height it is heading for. A hollow circle on a graph means "the curve comes right up to here, but this point is not included"; a filled circle means it is.

As x grows without bound. A limit can also describe what happens as x → ∞. For g(x) = (2x + 1) ÷ (x − 3):

so lim_x→∞ g(x) = 2. That is exactly the horizontal asymptote y = 2 you met in 2.8: an asymptote is a limit drawn as a line.

4The gradient of a curve, as a limit

A straight line has the same gradient everywhere. A curve does not: y = x² is flat at the origin and steep further out. So "the gradient of the curve at P" needs a definition, and the definition is this.

The gradient of a curve at a point P is the gradient of the tangent at P, the straight line that touches the curve at P and runs in the same direction as the curve there.

You cannot measure a tangent's gradient with rise over run directly, because you only know one point on it. So approximate. Take a second point Q on the curve, join P to Q with a chord, and find the chord's gradient. Then move Q closer to P and do it again.

For y = x² and P(3, 9), let Q be the point with x-coordinate 3 + h, where h is a small step.

As h → 0 the chord gradients approach 6, from above when Q is to the right and from below when it is to the left. So the gradient of y = x² at (3, 9) is 6. Figure 3 shows the chords swinging round to the tangent as Q slides down to P.

Notice what the limit did. At h = 0 the chord gradient would be 0 ÷ 0, meaningless, because P and Q would be the same point. The limit asks what happens near h = 0 instead, which is exactly what section 2 was for. This is why limits come first in calculus.

Why a tangent makes sense at all. Zoom in far enough on any smooth curve and it looks like a straight line. Figure 4 does it for y = x² at (1, 1): by the third window the curve is indistinguishable from a line of gradient 2. The gradient of the curve at a point is the gradient of that line you see when you zoom in.

Your GDC does precisely this when it reports a gradient: it takes a very small h and calculates the chord gradient. That is why its answers are occasionally a hair away from the exact value, 5.9999999 instead of 6.

5The derivative as the gradient function

Repeat section 4 at other points of y = x². The same algebra, with a general point (x, x²) in place of (3, 9), gives a chord gradient of 2x + h, which tends to 2x. So:

Every gradient is double the x-coordinate. The gradient is itself a function of x, and Figure 5 plots it.

That new function is called the derivative, or the gradient function, and finding it is called differentiation. For f(x) = x², the derivative is f′(x) = 2x. Put in any x and it hands you the gradient of the original curve at that x.

Read Figure 5 panel by panel, because the exam will ask you to move between the two graphs.

• Where f is falling (x < 0), its tangents slope down, so f′(x) is negative: panel (b) is below the axis.
• At the bottom of the bowl the tangent is horizontal, so f′(0) = 0: panel (b) crosses the axis.
• Where f is rising (x > 0), f′(x) is positive, and it grows as the curve steepens.

Subtopic 5.2 turns those three bullet points into a method, and 5.3 gives you the rule that produced 2x without any tables.

6The derivative as a rate of change, and its notation

A gradient is a rate of change: how fast y changes compared with x. For a straight line the rate is constant (2.1). For a curve it varies, and the derivative gives the rate at one precise value, the instantaneous rate of change.

The chord from section 4 has a meaning too. Its gradient is the average rate of change over the interval from P to Q. The derivative is what the average rate tends to as the interval shrinks to nothing.

A ball is thrown upwards, and its height after t seconds is h(t) = 20t − 5t² metres. Figure 6 draws it.

The ball is slowing down all the time, so at t = 1 exactly it must be going faster than its average over the next second. Shrink the interval:

The averages tend to 10, so at the instant t = 1 the ball is rising at 10 m s⁻¹. That is the gradient of the tangent in Figure 6, and it is the derivative of h at t = 1.

The notation. The guide expects you to read and write all of these. They mean the same kind of thing: "the derivative of the top quantity with respect to the bottom one".

dy/dx comes from the Δy/Δx of a chord, the Greek letter for "a change in", with the d marking that the change has shrunk to the limit. It is not a fraction you can cancel: the d's do not divide out. Treat dy/dx as one symbol.

The value at a point is written either f′(1) = 10 or dh/dt = 10 when t = 1. Both are fine.

Units and interpretation. The units of a derivative are always units of the top per unit of the bottom. If V is in cm³ and r in cm, dV/dr is in cm³ per cm. If s is in metres and t in seconds, ds/dt is in m s⁻¹. This gives you the template for every "interpret" question:

dy/dx = k when x = a means: at the instant x = a, y is increasing at k (units of y) per (unit of x). A negative k means y is decreasing.

For a sphere, V = (4/3)πr³. The chord method at r = 3 cm gives

so dV/dr ≈ 113 cm³ per cm when r = 3. In words: when the radius is 3 cm, the volume is increasing at about 113 cm³ for each extra centimetre of radius. (The exact value is 36π, which 5.3 will let you find in one line.)

7Estimating a derivative with technology

On Paper 2 you are expected to use technology. Every GDC has a numerical derivative, usually written d/dx or reached from the graph screen as dy/dx at a chosen x. It works by the chord method with a tiny h. Two things to remember:

• Round sensibly. If the screen says 9.0000003, the derivative is 9. Give other answers to 3 significant figures unless told otherwise.
• The GDC gives the gradient at one point, not the gradient function. When a question wants f′(x) as an expression, you must find it algebraically (5.3 and 5.6); when it wants f′(2) as a number on Paper 2, the GDC is enough, and you should write "using GDC" beside the value.

For a table of chord gradients, as in section 4, type the chord-gradient expression into the GDC's table and let it fill in the values. The working mark is for showing the expression you used.