Differentiation Revision Notes — 9709 Paper 1

What the derivative is

Syllabus requirement · §1.7.1

“

understand the gradient of a curve at a point as the limit of the gradients of a suitable sequence of chords, and use the notations f′(x), f″(x), dy/dx and d²y/dx² for first and second derivatives

”

The problem this solves

A straight line has one gradient, and you find it from any two points on it:

m = \frac{y_2 - y_1}{x_2 - x_1}

That formula needs two points. A curve does not have one gradient — it is steep in places and flat in others — so "the gradient of the curve at the single point $P$ " is not yet a thing you can compute. There is only one point, and the formula wants two.

Differentiation is the fix. The idea is to take the two-point formula, which you already trust, and squeeze the two points together until they are (almost) the same point.

Chords, and sliding one end in

Pick your point $P$ on the curve. Pick a second point $Q$ , further along the same curve. The straight line joining them is a chord, and its gradient is something you can work out, because you have two points.

That chord gradient is not the answer — a chord cuts across the curve, so it is too steep or too shallow. But now slide $Q$ along the curve towards $P$ . The chord swings round, and the closer $Q$ gets, the better the chord matches the direction the curve is actually heading in at $P$ .

The chord gradients do not wander about as you do this. They close in on one number. That number is what we call the gradient of the curve at $P$ , and the line through $P$ with that gradient is the tangent.

Each grey chord is a genuine gradient you could calculate from two points. As Q slides towards P, they close in on the red tangent — that limit is what dy/dx means.

The definition, in one sentence

The gradient of a curve at $P$ is the number that the gradients of the chords $PQ$ approach as $Q$ slides towards $P$ .

The word "approach" is doing real work: the chord gradient never actually equals the answer, because when $Q$ reaches $P$ you have one point again and the fraction becomes $\tfrac{0}{0}$ . It gets arbitrarily close, and the number it gets close to is the one we want. That is all "limit" means here.

Watch it happen with numbers

Nothing above is convincing until you see the numbers settle. Take the simplest curve, $y = x^2$ , and the point $P(1, 1)$ on it.

For a second point $Q$ with $x$ -coordinate $q$ , the $y$ -coordinate is $q^2$ , so

\text{gradient of } PQ = \frac{q^2 - 1}{q - 1}

Now put in values of $q$ closing in on $1$ from above:

$q$	$Q = (q,\ q^2)$	Gradient of chord $PQ$
$2$	$(2,\ 4)$	$\dfrac{4 - 1}{2 - 1} = 3$
$1.5$	$(1.5,\ 2.25)$	$\dfrac{2.25 - 1}{0.5} = 2.5$
$1.1$	$(1.1,\ 1.21)$	$\dfrac{1.21 - 1}{0.1} = 2.1$
$1.01$	$(1.01,\ 1.0201)$	$\dfrac{0.0201}{0.01} = 2.01$
$1.001$	$(1.001,\ 1.002001)$	$\dfrac{0.002001}{0.001} = 2.001$

The gradients are not drifting anywhere — they are marching towards 2.

Read the right-hand column downwards: $3,\ 2.5,\ 2.1,\ 2.01,\ 2.001$ . There is no mystery about where this is going. The gradient of $y = x^2$ at the point $(1,1)$ is $\mathbf{2}$ .

And when you learn the power rule in the next section you will get $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x$ , which at $x = 1$ gives $2$ . The table and the rule agree, because the rule was built from exactly this process.

The same thing with algebra instead of numbers

Decimals get you close but never all the way. Algebra gets you there exactly, and this is the version the examiners ask for.

Instead of naming the second point $1.001$ , call its $x$ -coordinate $1 + h$ , where $h$ is a small step whose size we have not fixed. Then:

Q = \left(1 + h,\ (1+h)^2\right)

\text{gradient of } PQ = \frac{(1+h)^2 - 1}{(1+h) - 1}

Expand the top: $(1+h)^2 = 1 + 2h + h^2$ , so the top is $1 + 2h + h^2 - 1 = 2h + h^2$ . The bottom is just $h$ . So

\text{gradient of } PQ = \frac{2h + h^2}{h} = \frac{h(2 + h)}{h} = 2 + h

Now the limit is obvious rather than guessed. Every chord has gradient exactly $2 + h$ ; as $h \to 0$ that becomes $2$ . Same answer as the table, with no rounding anywhere.

Why you may cancel the h

Cancelling $\dfrac{h(2+h)}{h}$ to $2 + h$ is legal because $h \neq 0$ — the two points are genuinely distinct while the chord exists. You only let $h$ go to zero after the cancelling, which is precisely the trick that dodges $\tfrac{0}{0}$ .

The chord question, as the exam sets it

9709/12 O/N 2024 Q35 marks

The equation of a curve is $y = 2x^2 - 3$ . Two points $A$ and $B$ with $x$ -coordinates $2$ and $(2+h)$ respectively lie on the curve.

(a) Find and simplify an expression for the gradient of the chord $AB$ in terms of $h$ . [3]

(b) Explain how the gradient of the curve at the point $A$ can be deduced from the answer to part (a), and state the value of this gradient. [2]

Show full working

1
(a) Start by finding the two $y$ -coordinates. The curve is $y = 2x^2 - 3$ , so at $x = 2$ : $y_A = 2(2)^2 - 3 = 2 \times 4 - 3 = 8 - 3 = 5$ So $A = (2,\ 5)$ .
Do this first and write it down. The mark scheme gives a mark for the y-coordinate at 2+h and expects your 5 to have come from 2(2)²−3.
2
At $x = 2 + h$ , substitute $2+h$ wherever $x$ appears: $y_B = 2(2+h)^2 - 3$ So $B = \left(2+h,\ 2(2+h)^2 - 3\right)$ .
Leave it unexpanded for one line. Substituting is one idea; expanding is the next.
3
Now the ordinary two-point gradient formula, $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ , with $B$ as point 2 and $A$ as point 1: $m_{AB} = \frac{\left(2(2+h)^2 - 3\right) - 5}{(2+h) - 2}$
Nothing new has happened yet — this is the same formula you have used since coordinate geometry. Only the letters are unusual.
4
Deal with the bottom first, because it is easy and it tells you what has to cancel: $(2+h) - 2 = h$
The denominator is always h in these questions. Knowing that in advance tells you the numerator must end up with a factor of h.
5
Now the top. Expand the bracket: $(2+h)^2 = (2+h)(2+h) = 4 + 2h + 2h + h^2 = 4 + 4h + h^2$
Write out all four products. (2+h)² = 4 + h² is the classic slip in this question.
6
Multiply by $2$ : $2(2+h)^2 = 2\left(4 + 4h + h^2\right) = 8 + 8h + 2h^2$
7
Subtract the $3$ , then subtract the $5$ : $\left(8 + 8h + 2h^2\right) - 3 - 5 = 2h^2 + 8h + 0 = 2h^2 + 8h$
The constants cancel completely — 8 − 3 − 5 = 0. They always will, because A lies on the curve. If a constant survives, you have made an arithmetic error.
8
Put the top over the bottom: $m_{AB} = \frac{2h^2 + 8h}{h}$
9
Factorise the numerator so the cancelling is visible: $m_{AB} = \frac{h\left(2h + 8\right)}{h}$
10
Cancel the $h$ (legal, since $h \neq 0$ — $A$ and $B$ are different points): $m_{AB} = 2h + 8$
Answer to (a). The mark scheme accepts 2h + 8 or 2(h + 4).
11
(b) Now say what happens as $B$ slides down to $A$ . The $x$ -coordinate of $B$ is $2 + h$ , so $B$ reaching $A$ means $h \to 0$ , and the chord becomes the tangent.
The mark scheme wants one of two statements: “h → 0”, or “chord AB → tangent at A”. Write one of them explicitly — it is a mark on its own.
12
Put $h = 0$ into the expression from (a): $m_{AB} = 2(0) + 8 = 8$
Only now, after the cancelling, is putting h = 0 safe.
13
Looking ahead: the power rule in the next section gives $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x$ for this curve, and at $x = 2$ that is $4 \times 2 = 8$ — the same answer. Nothing here depends on that; it is simply where the shortcut comes from.
Once you have the power rule this becomes a five-second check on questions like this one. For now it is just reassurance that the two routes agree.

Answer

(a) $2h + 8$ (b) As $h \to 0$ the chord $AB$ approaches the tangent at $A$ , so the gradient at $A$ is $\mathbf{8}$

The structure never changes: form $\dfrac{\mathrm{f}(a+h) - \mathrm{f}(a)}{h}$ , expand, watch the constants cancel, factor out $h$ , cancel it, then set $h = 0$ . If the $h$ will not cancel, you have expanded something wrongly.

The other format: a table of chord gradients

The same idea gets tested numerically, with the arithmetic done for you and a table of chords closing in on a point. You are asked what the table suggests. The answer is always "the number the column is heading towards".

Reading the limit off a table

9709/13 O/N 2025 Q33 marks

The equation of a curve is $y = \mathrm{f}(x)$ , where $\mathrm{f}(x) = \tfrac12 x^{\frac23}(x-2)^2$ . The following points lie on the curve; non-exact $y$ -coordinates are given correct to 6 decimal places.

A(8,\ 72), \quad B(8.001,\ k), \quad C(8.01,\ 72.300388), \quad D(8.1,\ 75.038882)

(a) Find the value of $k$ , correct to 6 decimal places. [1]

(b) The table shows the gradients of chords $AB$ and $AC$ to 4 decimal places. Find the gradient of chord $AD$ , correct to 4 decimal places. [1]

Chord	$AB$	$AC$	$AD$
Gradient	$30.0039$	$30.0388$

(c) State what the values in the table suggest about the value of $\mathrm{f}'(8)$ . [1]

Show full working

1
(a) $k$ is simply $\mathrm{f}(8.001)$ — $B$ is a point on the curve, so its $y$ -coordinate comes from the formula. Substitute $x = 8.001$ : $\mathrm{f}(8.001) = \tfrac12 \times (8.001)^{\frac23} \times (8.001 - 2)^2$
2
Work the two pieces separately on the calculator. First the bracket: $8.001 - 2 = 6.001, \qquad (6.001)^2 = 36.012001$
3
Then the fractional power. Remember $x^{2/3}$ means "cube root, then square": $(8.001)^{\frac23} = \left(\sqrt[3]{8.001}\right)^2 = (2.0000833\ldots)^2 = 4.0003333\ldots$
Keep every digit your calculator shows. Rounding here destroys the 6th decimal place you are asked for.
4
Multiply the three parts: $\tfrac12 \times 4.0003333\ldots \times 36.012001 = 72.030004$
The mark scheme says CAO, “not AWRT” — 72.03 or 72.0300 would score zero. Give exactly 6 decimal places.
5
(b) A chord gradient is the ordinary two-point formula. Chord $AD$ joins $A(8, 72)$ to $D(8.1,\ 75.038882)$ : $m_{AD} = \frac{75.038882 - 72}{8.1 - 8}$
6
Top: $75.038882 - 72 = 3.038882$ . Bottom: $8.1 - 8 = 0.1$ . $m_{AD} = \frac{3.038882}{0.1} = 30.38882$
7
To 4 decimal places: $m_{AD} = 30.3888$
Dividing by 0.1 just moves the point one place right. The commonest error is dividing by 0.01 or by 8.1.
8
(c) Now line the three chord gradients up in order of how close the second point is to $A$ : $\underbrace{30.3888}_{AD,\ \text{step } 0.1}, \qquad \underbrace{30.0388}_{AC,\ \text{step } 0.01}, \qquad \underbrace{30.0039}_{AB,\ \text{step } 0.001}$
Sort them by step size, not by the order printed in the table. The pattern only shows up in this order.
9
Each time the step shrinks by a factor of $10$ , the gradient loses a digit of its excess over $30$ : $0.3888$ , then $0.0388$ , then $0.0039$ . The values are closing in on $30$ . $\mathrm{f}'(8) \approx 30$
The mark scheme wants “30”, optionally hedged with “approximately”. Do not write 30.0039 — that is a chord gradient, not the tangent gradient.
10
(For interest: differentiating properly gives exactly $\mathrm{f}'(8) = 30$ , so the table is telling the truth.)

Answer

(a) $k = 72.030004$ (b) $30.3888$ (c) $\mathrm{f}'(8) \approx 30$

These parts are one mark each and pure calculator work — never leave them blank. The only skills are: substitute into the given formula, use $\dfrac{y_2-y_1}{x_2-x_1}$ , and say the round number the column is approaching.

The four notations

Once the gradient function exists it needs a name, and the syllabus expects you to recognise four.

Notation	Read as	Used when
$\dfrac{\mathrm{d}y}{\mathrm{d}x}$	dee y by dee x	the curve is given as $y = \ldots$
$\mathrm{f}'(x)$	f dashed of x	the curve is given as $\mathrm{f}(x) = \ldots$
$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$	dee two y by dee x squared	testing the nature of a stationary point
$\mathrm{f}''(x)$	f double-dashed of x	the same, in function notation

The syllabus expects all four. Answer in whichever notation the question used.

dy/dx is not a fraction

Despite how it looks, $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ is one symbol meaning "the derivative of $y$ with respect to $x$ ". You cannot cancel the $\mathrm{d}$ 's, and $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ is not $\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2$ .

The one place it behaves like a fraction is the chain rule (next section, and again in Connected rates of change), where the middle letters do cancel. That is a genuine result, not a licence to cancel elsewhere.

What the second derivative means

Differentiate once and you get the gradient function. Differentiate that, and you get the rate at which the gradient itself is changing:

y \ \xrightarrow{\ \text{differentiate}\ }\ \frac{\mathrm{d}y}{\mathrm{d}x} \ \xrightarrow{\ \text{differentiate}\ }\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2}

$\dfrac{\mathrm{d}y}{\mathrm{d}x}$ answers "is the curve going up or down, and how steeply?"
$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ answers "is that steepness increasing or decreasing?"

If $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0$ the gradient is getting bigger as you move right — the curve bends upwards, like a $\cup$ . If it is negative the gradient is falling — the curve bends downwards, like a $\cap$ . Hold on to that picture; it is the whole reason the second-derivative test in the Stationary points section works.

Everything in this section is about what a derivative means. Actually producing one needs a rule, and that is the next section — so the worked examples above use nothing but the two-point gradient formula and algebra.

What you are not asked for

Formal differentiation from first principles is not required on Paper 1, and neither is any formal treatment of limits — the syllabus says only an informal understanding is expected. The chord picture, the table of values and the $h \to 0$ argument above are exactly the level asked for. You will never be told to "differentiate from first principles".

In the exam

17 parts · 29 marks · in 9 of 37 papers, 2021–2025

Questions on this idea are short, 1–3 marks a part, and they come in the two layouts shown above: a chord from $x = a$ to $x = a + h$ worked in algebra, or a table of chord gradients worked on the calculator. Both are pure method. A student who has practised each layout once will not be surprised by either.

Your turn

The first builds the limit numerically, the second does it algebraically, the third is a past-paper table. Do not use the power rule in questions 1 and 2 — the point is the process it came from.

1
The point $P(3, 9)$ lies on the curve $y = x^2$ .

(a) Find the gradient of the chord joining $P$ to the point $Q$ on the curve with $x$ -coordinate $3.1$ , and then to the point with $x$ -coordinate $3.01$ .

(b) What do these suggest the gradient of the curve at $P$ is?
Stuck? Show hint
For each $Q$ , work out its $y$ -coordinate from $y = x^2$ first, then use $\dfrac{y_2 - y_1}{x_2 - x_1}$ . Keep all decimal places.
Show solution
1. 1
  (a) First chord. At $x = 3.1$ the $y$ -coordinate is $y = (3.1)^2 = 9.61$ so $Q = (3.1,\ 9.61)$ .
2. 2
  Gradient of $PQ$ : $m = \frac{9.61 - 9}{3.1 - 3} = \frac{0.61}{0.1} = 6.1$
  Subtract the y's, subtract the x's, divide. Nothing here is new — it is the gradient formula from coordinate geometry.
3. 3
  Second chord. At $x = 3.01$ : $y = (3.01)^2 = 9.0601$ so the second point is $(3.01,\ 9.0601)$ .
4. 4
  Gradient: $m = \frac{9.0601 - 9}{3.01 - 3} = \frac{0.0601}{0.01} = 6.01$
5. 5
  (b) The two gradients are $6.1$ and $6.01$ . As the second point gets ten times closer, the gradient gets ten times closer to $6$ .
  State the pattern, not just the number — that is what “what do these suggest” is asking for.
6. 6
  So the gradient of the curve at $P$ is $6$ .
  After the next section you will be able to confirm this in one line — y = x² gives dy/dx = 2x, and 2 × 3 = 6. The chords got there first, and without any rule.
Answer
(a) $6.1$ and $6.01$ (b) The gradient at $P$ is $6$
25 marks
The points $A$ and $B$ lie on the curve $y = x^2 + 5x$ , with $x$ -coordinates $1$ and $1 + h$ respectively.

(a) Show that the gradient of the chord $AB$ is $7 + h$ .

(b) Hence write down the gradient of the curve at $A$ .
Stuck? Show hint
Find $y_A$ from the equation, then $y_B$ by substituting $1+h$ for $x$ . Expand $(1+h)^2$ fully, and expect every constant to cancel.
Show solution
1. 1
  (a) The $y$ -coordinate at $x = 1$ : $y_A = (1)^2 + 5(1) = 1 + 5 = 6$ so $A = (1,\ 6)$ .
2. 2
  The $y$ -coordinate at $x = 1 + h$ — substitute $1+h$ everywhere $x$ appears: $y_B = (1+h)^2 + 5(1+h)$
3. 3
  Expand the square: $(1+h)^2 = (1+h)(1+h) = 1 + h + h + h^2 = 1 + 2h + h^2$
  Four products, written out. (1+h)² = 1 + h² loses the whole question.
4. 4
  Expand the other bracket: $5(1+h) = 5 + 5h$
5. 5
  Add them: $y_B = \left(1 + 2h + h^2\right) + \left(5 + 5h\right) = h^2 + 7h + 6$
  Collect like terms carefully: 2h + 5h = 7h, and 1 + 5 = 6.
6. 6
  Now the gradient formula. The bottom first: $(1+h) - 1 = h$
7. 7
  The top: $y_B - y_A = \left(h^2 + 7h + 6\right) - 6 = h^2 + 7h$
  The constant 6 cancels — as it must, since A is on the curve.
8. 8
  So $m_{AB} = \frac{h^2 + 7h}{h}$
9. 9
  Factorise the top to expose the common factor: $m_{AB} = \frac{h\left(h + 7\right)}{h}$
10. 10
  Cancel $h$ , which is allowed because $h \neq 0$ : $m_{AB} = h + 7 = 7 + h$ as required.
11. 11
  (b) As $B$ slides towards $A$ , $h \to 0$ and the chord becomes the tangent: $m = 7 + 0 = 7$
  Say the words “as h → 0” — in the real mark scheme that statement carries its own mark.
Answer
(a) shown (b) Gradient at $A$ is $7$
39709/11 M/J 2024 Q43 marks
The equation of a curve is $y = \mathrm{f}(x)$ , where $\mathrm{f}(x) = (2x-1)\sqrt{3x-2} - 2$ . The following points lie on the curve. Non-exact values have been given correct to 5 decimal places.

$A(2,\ 4)$ , $B(2.0001,\ k)$ , $C(2.001,\ 4.00625)$ , $D(2.01,\ 4.06261)$ , $E(2.1,\ 4.63566)$ , $F(3,\ 11.22876)$

(a) Find the value of $k$ . Give your answer correct to 5 decimal places. [1]

(b) The table shows the gradients of the chords $AB$ , $AC$ , $AD$ and $AF$ .

Chord $AB$ $AC$ $AD$ $AE$ $AF$
Gradient of chord $6.2501$ $6.2511$ $6.2608$ $7.2288$

Find the gradient of the chord $AE$ . Give your answer correct to 4 decimal places. [1]

(c) Deduce the value of $\mathrm{f}'(2)$ using the values in the table. [1]
Stuck? Show hint
(a) is $\mathrm{f}(2.0001)$ on the calculator. (b) is the two-point gradient formula. For (c), read the gradients in order of how close the second point is to $A$ .
Show solution
1. 1
  (a) $B$ lies on the curve, so $k = \mathrm{f}(2.0001)$ . Work the pieces separately: $2(2.0001) - 1 = 3.0002, \qquad 3(2.0001) - 2 = 4.0003$
2. 2
  Then $k = 3.0002 \times \sqrt{4.0003} - 2 = 3.0002 \times 2.0000750 \ldots - 2 = 4.000625\ldots$
  Keep every digit on the calculator until the end.
3. 3
  To 5 decimal places: $k = 4.00063$
  The mark scheme is CAO: 4.0006 (only 4 decimal places) scores nothing.
4. 4
  (b) Chord $AE$ joins $A(2, 4)$ to $E(2.1,\ 4.63566)$ : $m_{AE} = \frac{4.63566 - 4}{2.1 - 2} = \frac{0.63566}{0.1} = 6.3566$
5. 5
  (c) Put the chords in order of how close the second point is to $A$ , furthest first: $AF: 7.2288, \quad AE: 6.3566, \quad AD: 6.2608, \quad AC: 6.2511, \quad AB: 6.2501$
  Sort by the size of the step (1, 0.1, 0.01, 0.001, 0.0001), not by the order of the table.
6. 6
  Each time the step shrinks by a factor of $10$ , the excess over $6.25$ shrinks by about $10$ too: $0.1066$ , $0.0108$ , $0.0011$ , $0.0001$ . The gradients are closing in on $6.25$ . $\mathrm{f}'(2) = 6.25$
  The mark scheme is CAO on 6.25. Quoting 6.2501, the gradient of the closest chord, is not the limit.
Answer
(a) $k = 4.00063$ (b) $6.3566$ (c) $\mathrm{f}'(2) = 6.25$

Chord	$AB$	$AC$	$AD$	$AE$	$AF$
Gradient of chord	$6.2501$	$6.2511$	$6.2608$		$7.2288$

Checking your access…

Can you do all of these?

Explain what $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ means using chords and a limit
Find the gradient of the chord from $x = a$ to $x = a+h$ , simplify it, and let $h \to 0$
Read the value of $\mathrm{f}'(a)$ off a table of chord gradients
Rewrite $\dfrac{3}{x^2}$ , $5\sqrt{x}$ , $(2x-1)(x+3)$ and $\dfrac{x^3+2x}{x}$ ready for differentiating
Differentiate $(3x+2)^{1/2}$ , $\dfrac{4}{(2x-1)^2}$ and $\dfrac{3}{2x^2-5}$ with the chain rule
Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ and $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ for an equation containing an unknown constant
Find the tangent and the normal at a point, and give the answer as $ax+by+c=0$ with integers
Find where a curve has a stated gradient, and handle " $y = mx+k$ is a tangent to the curve"
Use $b^2 - 4ac = 0$ when a line with an unknown gradient is a tangent
Find the set of values of $x$ for which a given curve is decreasing — with "and"/"or" written correctly, and any excluded $x$ left out
Justify "increasing, decreasing or neither" from the sign of $\mathrm{f}'(x)$ — term by term or by completing the square — not from sample values
Find both stationary points of a cubic and determine their nature with $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$
Say why a stationary point with $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0$ is a minimum
Sketch a curve from its stationary points, and say where they go under a translation or a reflection
Find an unknown constant, or two, in a curve's equation from a given stationary point
Use a constraint to write a volume or area in one variable, then maximise it
Do a connected-rates problem in either direction, and give the answer with units
Recover a curve from $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ , using the stationary point for $c_1$ and the point for $c_2$