Differentiation

Syllabus coverage

Extended only: estimate gradients of curves by drawing tangents
Extended only: the $\frac{d y}{d x}$ notation will be expected
Extended only: use the derivatives of functions of the form $a x^{n}$ , where a is a rational constant and n is a positive integer or zero, and simple sums of not more than three of these
Extended only: apply differentiation to gradients and stationary points (turning points)
Extended only: discriminate between maxima and minima by any method - an accurate sketch, the second differential, or the gradient either side of the turning point
Extended only: candidates are not expected to identify points of inflection
The rate-of-change idea is applied to simple kinematics (the graphs in practical situations topic), so a displacement function can be differentiated for velocity and acceleration

The gradient of a curve at a point is the gradient of the tangent at that point. Drawing the tangent estimates the gradient; differentiating gives it exactly.

The gradient of a curve is different at every point

The gradient of a straight line is the same all the way along it: rise divided by run, and it does not matter which two points on the line you choose. A curve is not like that. Take $y = x^{2}$ : at $(0, 0)$ it is flat, at $(1, 1)$ it is climbing gently, and at $(3, 9)$ it is climbing steeply. The gradient of a curve only means anything at a single point, and it is defined as the gradient of the straight line that just touches the curve there.

That line is the tangent. Cambridge's own wording is that a tangent should touch the curve at the required point and stay in contact with the curve for the minimum possible distance, and that it should not cross the curve at that point. So a tangent grazes the curve; it is not a chord that cuts through it at two points.

To estimate a gradient by drawing: mark the point, lay a ruler so the line touches the curve only there, draw the tangent long enough to reach the edges of the grid, then pick two points on your tangent as far apart as possible and work out rise over run. The further apart those two points, the smaller the percentage error in your reading, because a small slip in reading a coordinate matters less over a long run.

Gradient of a straight line: rise divided by run, the same everywhere on the line
Gradient of a curve: the gradient of the tangent at that one point, and it changes as you move along the curve
A positive gradient means the curve is rising, a negative gradient means it is falling, and zero means the tangent is horizontal
An estimate from a drawn tangent is always approximate, so read the two points from the grid as accurately as you can

Cambridge expects tangents drawn with a ruler, touching the curve at the point and not crossing it. A tangent that cuts through the curve can lose the accuracy mark even when the arithmetic afterwards is perfect.

dy/dx: what the derivative means

A tangent is drawn by eye, so it can only ever give an estimate. The derivative gives the exact gradient, and the notation Cambridge expects is $\frac{d y}{d x}$ . Read it as 'the rate of change of y with respect to x', or more simply as 'the gradient of the curve'.

The idea behind the notation is a chord. Take a point P on the curve and a second point Q a little way along it. The gradient of the chord PQ is $\frac{δ y}{δ x}$ , where $δ y$ and $δ x$ are the small changes in y and x between P and Q. Now slide Q closer and closer to P. The chord pivots about P, gets closer and closer to the tangent, and $\frac{δ y}{δ x}$ gets closer and closer to the gradient of that tangent. The value it settles on as $δ x$ tends to zero is written $\frac{d y}{d x}$ , and that is the whole meaning of the symbol.

You never work that out by hand. There is a rule that turns the idea into one line of algebra, and it works for every power of x you meet in 0580. The table below shows the chord gradients for $y = x^{2}$ at $x = 3$ as the second point closes in.

Value of $δ x$	Gradient of the chord $\frac{δ y}{δ x}$
1	7
0.5	6.5
0.1	6.1
0.01	6.01
tending to 0	tending to 6

For

y = x^{2}

x = 3

the chord gradient is always

6 + δ x

. As

δ x

shrinks, the answer closes in on 6, and differentiating gives

\frac{d y}{d x} = 2 x = 6

x = 3

. A good estimate from a tangent and the exact derivative should agree to about one decimal place.

Differentiating powers of x

The rule is short: if $y = a x^{n}$ then $\frac{d y}{d x} = n a x^{n - 1}$ . Multiply by the index, then reduce the index by one. It applies to each term of a sum separately, because the gradient of a sum is the sum of the gradients of its parts.

Two cases catch people out. A term with no x in it at all is a constant, and a constant is a horizontal line with gradient zero, so it differentiates to 0. A term such as $7 x$ is really $7 x^{1}$ , so the index is 1 and $1 \times 7 x^{0} = 7$ .

The syllabus asks for functions of the form $a x^{n}$ , where a is a rational constant and n is a positive integer or zero, and for simple sums of not more than three of these. So $y = 5 x^{4} - 3 x^{2} + 8$ is exactly the kind of thing you will be asked, and three terms is the most you should expect in one expression.

$y$	$\frac{d y}{d x}$	Why
$x^{5}$	$5 x^{4}$	multiply by the index 5, then take 1 off it
$4 x^{3}$	$12 x^{2}$	$3 \times 4 = 12$ , and the index becomes 2
$7 x$	$7$	$7 x = 7 x^{1}$ , so $1 \times 7 x^{0} = 7$
$12$	$0$	a constant is a horizontal line, gradient 0
$5 x^{4} - 3 x^{2} + 8$	$20 x^{3} - 6 x$	differentiate each term; the 8 gives 0

In 0580 the index n is a positive integer or zero, so the power of x never goes below zero on this course. Negative and fractional indices are not examined in this syllabus and a 0580 question will not ask you to differentiate

x^{- 2}

\sqrt{x}

The gradient at a point, and the equation of a tangent

Differentiating gives you a formula for the gradient, not a single number. To get the gradient at one particular point, differentiate and then substitute the x-coordinate of that point. For $y = x^{2} - 4 x + 7$ , $\frac{d y}{d x} = 2 x - 4$ , so at $x = 3$ the gradient is $2 (3) - 4 = 2$ .

The tangent at that point is the straight line through the point with that gradient. Use the coordinate geometry form $y - y_{1} = m (x - x_{1})$ , where $(x_{1}, y_{1})$ is the point on the curve and m is the gradient you have just found. You need the y-coordinate, and it comes from the original equation of the curve, never from the derivative.

The normal is the line through the same point perpendicular to the tangent. Perpendicular gradients multiply to give $- 1$ , so if the tangent has gradient m then the normal has gradient $- \frac{1}{m}$ . Finding the normal is therefore the same calculation with one extra line.

Step	What to do	For $y = x^{2} - 4 x + 7$ at $x = 3$
1	Differentiate	$\frac{d y}{d x} = 2 x - 4$
2	Substitute the x-value into the derivative: this is the gradient m	$m = 2 (3) - 4 = 2$
3	Substitute the same x-value into the original equation: this is $y_{1}$	$y_{1} = 9 - 12 + 7 = 4$
4	Tangent: $y - y_{1} = m (x - x_{1})$	$y - 4 = 2 (x - 3)$ , so $y = 2 x - 2$
5	Normal: gradient $- \frac{1}{m}$	$- \frac{1}{2}$ , so $y = - \frac{1}{2} x + \frac{11}{2}$

If the tangent is horizontal then m is 0 and the normal is vertical. A vertical line has no gradient, so the rule

m_{1} m_{2} = - 1

cannot be used there; write the normal as

x = a

, where a is the x-coordinate of the point.

Stationary points

A stationary point is a point on a curve where the gradient is zero, so the tangent there is horizontal. Cambridge also calls them turning points. They are the places where the curve stops rising and starts falling, or stops falling and starts rising.

To find them: differentiate, set $\frac{d y}{d x} = 0$ , solve for x, then substitute each x back into the original equation of the curve to get the matching y-coordinate. The derivative is always one degree lower than the curve, so a cubic curve has a quadratic derivative and can have up to two stationary points.

For $y = 2 x^{3} - 9 x^{2} + 12 x$ : $\frac{d y}{d x} = 6 x^{2} - 18 x + 12 = 6 (x - 1) (x - 2)$ , so $\frac{d y}{d x} = 0$ when $x = 1$ or $x = 2$ . Substituting into the curve gives $y = 2 - 9 + 12 = 5$ at $x = 1$ , and $y = 16 - 36 + 24 = 4$ at $x = 2$ . The stationary points are $(1, 5)$ and $(2, 4)$ .

Stationary point: a point where $\frac{d y}{d x} = 0$ and the tangent is horizontal
Solve $\frac{d y}{d x} = 0$ for x, then put each x back into the original equation to get y
Factorising the derivative is usually the quickest way to solve it, especially when every term shares a common factor
A stationary point is not automatically a maximum or a minimum, so read on to the next block

Never substitute the x-value into the derivative to find the y-coordinate. The derivative returns gradients, so at a stationary point it returns 0 for every curve.

Maximum or minimum?

A stationary point can be a maximum, where the curve was rising and now falls; a minimum, where it was falling and now rises; or neither. Cambridge accepts any method, and you should use whichever one the question makes quickest.

Method 1: the second derivative. Differentiate a second time to get $\frac{d^{2} y}{d x^{2}}$ , then substitute the x-value of the stationary point. A negative value means a maximum, because the gradient is going from positive towards negative. A positive value means a minimum. For $y = 2 x^{3} - 9 x^{2} + 12 x$ , $\frac{d^{2} y}{d x^{2}} = 12 x - 18$ ; at $x = 1$ that is $- 6$ , so $(1, 5)$ is a maximum, and at $x = 2$ it is $6$ , so $(2, 4)$ is a minimum.

Method 2: the gradient either side. Work out $\frac{d y}{d x}$ at an x-value just below the stationary point and at one just above it. A gradient that goes from positive to negative means a maximum; from negative to positive means a minimum. For the same curve, $\frac{d y}{d x}$ is $0.66$ at $x = 0.9$ and $- 0.54$ at $x = 1.1$ , so $x = 1$ is a maximum.

Method 3: an accurate sketch. If you can see the shape of the curve, you can see which way it turns at each stationary point. This is the fastest method and the easiest to get wrong, so use it only when the shape is obvious.

Test	Maximum	Minimum
$\frac{d^{2} y}{d x^{2}}$ at the point	negative	positive
Gradient just before the point	positive	negative
Gradient just after the point	negative	positive
Shape of the curve there	bending downwards	bending upwards

\frac{d^{2} y}{d x^{2}} = 0

the second derivative test does not decide anything, so fall back on the gradient either side. The syllabus does not ask you to identify points of inflection, so every stationary point you are asked about is a maximum or a minimum.

Rate of change, and kinematics

$\frac{d y}{d x}$ is a rate of change: it says how fast y changes when x changes. That is why the same technique answers questions about speed and acceleration, and why the syllabus applies rate of change to simple kinematics.

If $s$ is the displacement of a particle from a fixed point, measured in metres, and $t$ is the time in seconds, then $\frac{d s}{d t}$ is the velocity in metres per second and $\frac{d^{2} s}{d t^{2}}$ is the acceleration in metres per second squared. The gradient of a displacement-time graph is the velocity, and the gradient of a velocity-time graph is the acceleration.

Take $s = t^{3} - 6 t^{2} + 9 t$ . Differentiating gives $v = \frac{d s}{d t} = 3 t^{2} - 12 t + 9$ , so at $t = 4$ the velocity is $3 (16) - 12 (4) + 9 = 9$ m/s. Differentiating again gives $a = \frac{d v}{d t} = 6 t - 12$ , so at $t = 4$ the acceleration is $24 - 12 = 12$ m/s $^{2}$ .

Displacement s in metres and time t in seconds: velocity $v = \frac{d s}{d t}$ and acceleration $a = \frac{d v}{d t} = \frac{d^{2} s}{d t^{2}}$
The particle is instantaneously at rest when $v = 0$ , which is a stationary point of the displacement function
A negative velocity is a real answer, not a mistake: the particle is moving in the negative sense of s

Units follow the letters. If s is in metres and t is in seconds, then

\frac{d s}{d t}

is in m/s and

\frac{d^{2} s}{d t^{2}}

is in m/s

^{2}

. Leave the units off and a correct number can still lose a mark.

Key terms you must be able to define

Gradient of a curve: The gradient of the tangent to the curve at that point. Unlike a straight line, a curve has a different gradient at each point.

Tangent: A straight line that touches a curve at a point and stays in contact with it for the minimum possible distance. It does not cross the curve at that point.

Chord: A straight line joining two points on a curve. The gradient of a chord estimates the gradient of the curve between those points.

$δ x$: A small change in the value of x. $\frac{δ y}{δ x}$ is the gradient of the chord joining two nearby points, and it tends to $\frac{d y}{d x}$ as $δ x$ tends to zero.

Derivative: The exact gradient of a curve, found by differentiating. It is a formula in x, not a single number.

$\frac{d y}{d x}$: The notation for the derivative: the rate of change of y with respect to x, that is the gradient of the curve.

Differentiation: The process of finding the derivative. For $y = a x^{n}$ the derivative is $n a x^{n - 1}$ : multiply by the index, then reduce the index by one.

Gradient function: The derivative thought of as a function: put in an x-value and it gives you the gradient of the curve at that point.

Stationary point: A point on a curve where $\frac{d y}{d x} = 0$ , so the tangent there is horizontal.

Turning point: Another name for a stationary point: a point where the curve turns from rising to falling, or from falling to rising.

Maximum point: A stationary point where the curve stops rising and starts falling. The second derivative is negative there.

Minimum point: A stationary point where the curve stops falling and starts rising. The second derivative is positive there.

Second derivative: The derivative of the derivative, written $\frac{d^{2} y}{d x^{2}}$ . Its sign at a stationary point decides whether the point is a maximum or a minimum.

Normal: The straight line through a point on a curve perpendicular to the tangent there. If the tangent has gradient m, the normal has gradient $- \frac{1}{m}$ .

Point of inflection: A point where a curve changes from bending one way to bending the other. 0580 does not ask you to identify these.

Rate of change: How fast one quantity changes as another changes. $\frac{d y}{d x}$ is the rate of change of y with respect to x.

Velocity: The rate of change of displacement: $v = \frac{d s}{d t}$ , in metres per second when s is in metres and t is in seconds.

Acceleration: The rate of change of velocity: $a = \frac{d v}{d t} = \frac{d^{2} s}{d t^{2}}$ , in metres per second squared.

In context

Cambridge sets 0580 in real situations - a recipe, a bank statement, a map, a garden, a graph of a car's journey. These show how the mathematics appears in a question, then you write your own working for the numbers you are given.

Estimating a gradient from a graph

The curve $y = x^{2}$ is drawn from $x = 0$ to $x = 4$ and you are asked to estimate the gradient at $(2, 4)$ . Draw the tangent there with a ruler, read two points on it as far apart as the grid allows, for example $(1, 0)$ and $(3, 8)$ , and work out $\frac{8 - 0}{3 - 1} = 4$ . Differentiating gives $\frac{d y}{d x} = 2 x$ , and at $x = 2$ that is exactly 4, so the estimate and the exact value agree.

Differentiating a sum of three terms

Differentiate $y = 3 x^{4} - 5 x^{2} + 2 x$ . Term by term: $3 x^{4}$ gives $12 x^{3}$ , $- 5 x^{2}$ gives $- 10 x$ , and $2 x$ gives 2, so $\frac{d y}{d x} = 12 x^{3} - 10 x + 2$ . Every term is differentiated, including the one whose index is 1.

Finding where the gradient has a given value

The curve $y = x^{3} - 3 x^{2} + 1$ has $\frac{d y}{d x} = 3 x^{2} - 6 x$ . To find where the gradient is 9, put $3 x^{2} - 6 x = 9$ , so $3 x^{2} - 6 x - 9 = 0$ , then $x^{2} - 2 x - 3 = 0$ , which factorises as $(x - 3) (x + 1) = 0$ . The gradient is 9 at $x = 3$ and at $x = - 1$ .

Tangent and normal

For $y = x^{2} - 4 x + 7$ at $x = 3$ , $\frac{d y}{d x} = 2 x - 4$ so the gradient is 2, and $y = 9 - 12 + 7 = 4$ , so the point is $(3, 4)$ . The tangent is $y - 4 = 2 (x - 3)$ , that is $y = 2 x - 2$ . The normal has gradient $- \frac{1}{2}$ , so $y - 4 = - \frac{1}{2} (x - 3)$ , that is $y = - \frac{1}{2} x + \frac{11}{2}$ .

Maximum or minimum

$y = 4 x - x^{2}$ , so $\frac{d y}{d x} = 4 - 2 x$ . Setting $\frac{d y}{d x} = 0$ gives $x = 2$ , and then $y = 8 - 4 = 4$ , so the only stationary point is $(2, 4)$ . The second derivative is $\frac{d^{2} y}{d x^{2}} = - 2$ , which is negative, so $(2, 4)$ is a maximum. The curve is an upside-down parabola, which agrees.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What does

\frac{d y}{d x}

mean?

The gradient of the curve at a point: the exact rate of change of y with respect to x.

Q2How do you estimate the gradient of a curve at a point?

Draw the tangent at that point with a ruler, read two points on the tangent as far apart as possible, and work out rise over run.

Q3Differentiate

y = x^{6}

\frac{d y}{d x} = 6 x^{5}

. Multiply by the index, then reduce the index by one.

Q4Differentiate

y = 9 x^{2} - 4 x + 7

\frac{d y}{d x} = 18 x - 4

. The 7 is a constant, so it differentiates to 0.

Q5What is the derivative of a constant?

0. A constant is a horizontal line, and a horizontal line has gradient zero.

Q6What is

\frac{d y}{d x}

for

y = 5 x

5. Write

5 x

5 x^{1}

, so the answer is

1 \times 5 x^{0} = 5

Q7How do you find the gradient of a curve at a given point?

Differentiate to get

\frac{d y}{d x}

, then substitute the x-coordinate of the point into it.

Q8How do you find the equation of the tangent to a curve at a point?

Get the gradient m from

\frac{d y}{d x}

, get the y-coordinate from the original equation, then use

y - y_{1} = m (x - x_{1})

Q9How are the gradients of a tangent and its normal related?

They are perpendicular, so their gradients multiply to

- 1

. The normal has gradient

- \frac{1}{m}

Q10How do you find the stationary points of a curve?

Set

\frac{d y}{d x} = 0

, solve for x, then substitute each x into the original equation to get the y-coordinate.

Q11What does the sign of the second derivative tell you at a stationary point?

Negative means a maximum; positive means a minimum.

Q12The second derivative at a stationary point is positive. What kind of point is it?

A minimum. The gradient is going from negative to positive, so the curve is bending upwards.

Q13How do you decide maximum or minimum without the second derivative?

Work out the gradient just either side of the point. Positive then negative is a maximum; negative then positive is a minimum.

Q14Displacement is

s = t^{3} - 6 t^{2} + 9 t

. What is the velocity?

v = \frac{d s}{d t} = 3 t^{2} - 12 t + 9

, in m/s when s is in metres and t is in seconds.

Q15How is acceleration found from a displacement function?

Differentiate twice:

a = \frac{d^{2} s}{d t^{2}}

, which is the derivative of the velocity.

0 of 15 marked as known

Exam technique for this topic

For an 'estimate the gradient' question, draw the tangent with a ruler so it touches the curve at the marked point and does not cross it, then read two points on the tangent as far apart as the grid allows. The mark is for a sensible tangent and a correct rise over run, not for a calculator value.
Always write the differentiation line down. A mark scheme awards a method mark for $\frac{d y}{d x}$ correct, and a candidate who jumps straight from the curve equation to the gradient at a point has no line to show for it.
When a question asks for the coordinates of a stationary point, give both numbers in brackets. Setting $\frac{d y}{d x} = 0$ produces the x-value only, and the y-value has to come from the original equation.
Use the second derivative to decide between a maximum and a minimum unless the question tells you which method to use. It is one extra line, it is unambiguous, and the sentence 'negative, so it is a maximum' picks up the reasoning mark as well as the answer.
Check the sense and the units of a kinematics answer. With s in metres and t in seconds the velocity is in m/s and the acceleration in m/s $^{2}$ , and a negative velocity is a perfectly good answer that tells you the particle is moving the other way.

Why students lose marks here

A student wroteThe derivative of

y = 4 x^{3} + 5

12 x^{2} + 5

Why it loses marksThe constant 5 was left unchanged. A constant is a horizontal line and a horizontal line has gradient zero, so the 5 differentiates to 0, not to itself.

The answer that scores

\frac{d y}{d x} = 12 x^{2}

. Differentiate each term separately:

4 x^{3}

gives

12 x^{2}

and the 5 gives 0.

A student wroteTo find the y-coordinate of a stationary point, substitute its x-value into

\frac{d y}{d x}

Why it loses marksThe derivative returns gradients, not points. At a stationary point

\frac{d y}{d x} = 0

, so this substitution gives 0 for every curve in the paper.

The answer that scoresSubstitute into the original equation. For

y = 2 x^{3} - 9 x^{2} + 12 x

x = 1

y = 2 - 9 + 12 = 5

, so the point is

(1, 5)

A student wroteThe second derivative at a stationary point is

- 6

, so the point is a minimum.

Why it loses marksThe sign has been read the wrong way round. A negative second derivative means the gradient is falling, so the curve is about to turn downwards: that is a maximum.

The answer that scoresNegative second derivative means a maximum, positive means a minimum.

- 6

x = 1

makes

(1, 5)

a maximum point.

A student wroteThe tangent to

y = x^{2} - 4 x + 7

x = 3

y = 2 x - 6

, using

y_{1} = 0

Why it loses marksThe y-coordinate was taken as zero instead of being worked out from the curve. The point where

x = 3

(3, 4)

, not

(3, 0)

, so the tangent line is shifted down by 4.

The answer that scores

y = 9 - 12 + 7 = 4

x = 3

, so

y - 4 = 2 (x - 3)

and the tangent is

y = 2 x - 2

Syllabus coverage

The gradient of a curve is different at every point

dy/dx: what the derivative means

Differentiating powers of x

The gradient at a point, and the equation of a tangent

Stationary points

Maximum or minimum?

Rate of change, and kinematics

Key terms you must be able to define

In context

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here