A graph rises, falls and rises again. A second graph describes how quickly the first graph changes. Can we use the second graph to reconstruct the first graph’s behaviour without seeing the original curve?
Yes—provided we understand what a derivative actually tells us.
This is one reason to study Additional Mathematics: a derivative is itself a function, and its graph can reveal where another function increases, decreases, turns or changes its manner of bending.
Students often learn to differentiate an equation and then substitute a number to find a gradient. That is an important skill. But a deeper understanding comes from reading the full derivative relationship rather than calculating one slope at a time.
A-Math offers students a language for connecting algebra, graphs and rates of change. This article develops that connection using precise examples, clear sign analysis and the distinction between what a derivative graph can tell us and what it cannot determine by itself.
Begin by separating two different graphs
Suppose a function is called f(x).
Its graph displays the value of f for each permitted input x.
The derivative function, written f'(x), describes the instantaneous gradient of f at each input where the derivative exists.
Its graph displays gradient values, not the original function’s heights.
For example, let:
f(x) = x².
Then:
f'(x) = 2x.
The graph of f is a parabola.
The graph of f’ is a straight line.
These graphs are different mathematical objects.
At x = 3, the original function has value 9, while its derivative has value 6.
The number 9 is the original output. The number 6 is its instantaneous rate of change per unit of x.
A student must keep those meanings separate.
Why the derivative graph can be easier to read
Consider the function f(x) = x² again.
Its derivative is f'(x) = 2x.
For negative x, the derivative is negative.
At x = 0, it is zero.
For positive x, the derivative is positive.
Therefore the original curve decreases as x approaches zero from the left, becomes stationary at zero and increases to the right.
The derivative graph tells us these facts through its position relative to the horizontal axis.
When f'(x) lies below zero, f is decreasing.
When f'(x) lies above zero, f is increasing.
Where f'(x) equals zero, f has a stationary candidate.
We can describe the curve’s general movement without relying solely on a sketch of f itself.
Positive derivative means increasing behaviour
For a differentiable function on an interval, a derivative that is strictly positive throughout that interval implies the function is strictly increasing there.
Consider:
f(x) = x³ + x.
Differentiate:
f'(x) = 3x² + 1.
Since x² is non-negative for every real x, the derivative is always greater than zero.
Therefore f increases throughout the real-number domain.
We have established this without plotting every input.
The derivative provides a general argument.
A student who simply tests a few values of x might suspect that the function rises everywhere, but the derivative shows why the conclusion holds for all real inputs.
That is an example of mathematical reasoning extending beyond numerical examples.
Negative derivative means decreasing behaviour
Now consider:
g(x) = -x³ – x.
Differentiate:
g'(x) = -3x² – 1.
For every real x, this derivative is negative.
Therefore g is strictly decreasing over all real inputs.
Its graph always moves downward as x increases.
Notice that negative derivative does not mean the original function value must be negative.
At x = -2:
g(-2) = 8 + 2 = 10.
The output is positive, but the derivative is negative.
The function can have a positive value while decreasing.
Students who confuse the sign of the output with the sign of its derivative can misread even simple graphs.
A complete cubic whose derivative is easy to factorise
Consider:
f(x) = 2x³ + 3x² – 12x + 1.
Differentiate:
f'(x) = 6x² + 6x – 12.
Factorise:
f'(x) = 6(x + 2)(x – 1).
The derivative is a quadratic.
It equals zero when:
x = -2
or:
x = 1.
These are stationary candidates of the original cubic.
The factorised derivative also tells us whether f is increasing or decreasing in the intervals separated by these values.
We do not yet need to calculate the original function’s output at each stationary input.
First, read the derivative signs.
Build a sign chart for the derivative
The critical inputs are -2 and 1.
For x < -2, both factors x + 2 and x – 1 are negative, so their product is positive.
Therefore f'(x) is positive.
For -2 < x < 1, the factor x + 2 is positive and x – 1 is negative.
Therefore f'(x) is negative.
For x > 1, both factors are positive.
Therefore f'(x) is positive.
The derivative-sign pattern is:
Positive, negative, positive.
This immediately tells us the original function:
Increases before x = -2.
Decreases between x = -2 and x = 1.
Increases after x = 1.
We have reconstructed the original curve’s direction changes from a quadratic derivative.
Classify the stationary points
At x = -2, the derivative changes from positive to negative.
The function rises before this point and falls afterward.
Therefore x = -2 corresponds to a local maximum.
At x = 1, the derivative changes from negative to positive.
The function falls before this point and rises afterward.
Therefore x = 1 corresponds to a local minimum.
These are local classifications.
A cubic with a positive leading coefficient increases without bound as x approaches positive infinity and decreases without bound towards negative infinity.
Therefore the local maximum is not a global maximum over the whole real line.
Similarly, the local minimum is not a global minimum over all real inputs.
The domain and precise language matter.
Find the coordinates on the original graph
The derivative graph gave us the stationary x-values and their nature.
To find the actual coordinates, return to the original function:
f(x) = 2x³ + 3x² – 12x + 1.
At x = -2:
f(-2) = 2(-8) + 3(4) – 12(-2) + 1.
This equals:
-16 + 12 + 24 + 1 = 21.
So the local maximum is (-2, 21).
At x = 1:
f(1) = 2 + 3 – 12 + 1 = -6.
So the local minimum is (1, -6).
The derivative alone did not provide those y-coordinates.
We needed the original function’s values as well.
This distinction becomes important when a question provides only a derivative graph.
Why the derivative graph cannot determine vertical position
Suppose we know:
f'(x) = 6(x + 2)(x – 1).
Integrating gives:
f(x) = 2x³ + 3x² – 12x + C,
where C is an arbitrary constant.
Every choice of C produces the same derivative graph.
The original cubic may be shifted up or down without changing any of its gradients.
Therefore the derivative graph by itself does not identify the absolute vertical position of f.
It can show increasing and decreasing intervals, stationary input values and local classifications.
But to recover a specific original function, we need additional information such as f(0) = 1.
That condition would determine C = 1.
This is the connection between a derivative graph and the constant of integration.
What the derivative graph looks like
For our cubic:
f'(x) = 6x² + 6x – 12.
The graph of f’ is an upward-opening parabola.
Its horizontal-axis intersections are at -2 and 1.
Between those roots, the derivative graph lies below the axis.
Outside them, it lies above.
These graph positions correspond exactly to the increasing and decreasing intervals of f.
The vertex of the derivative parabola is halfway between the two roots:
x = (-2 + 1)/2 = -1/2.
At that input, the derivative attains its smallest value.
Substitute:
f'(-1/2) = 6(3/2)(-3/2) = -27/2.
So the most negative slope of this cubic occurs at x = -1/2.
The original function is decreasing fastest with respect to x at that location.
What the minimum of f’ tells us about f
The derivative f'(x) is itself a function.
Its gradient is the second derivative f”(x).
For our example:
f”(x) = 12x + 6.
This equals zero at:
x = -1/2.
It is negative for x < -1/2 and positive for x > -1/2.
The original cubic changes concavity at x = -1/2.
Therefore it has a point of inflexion there.
Find the function value:
f(-1/2) = 2(-1/8) + 3(1/4) – 12(-1/2) + 1.
Simplify:
-1/4 + 3/4 + 6 + 1 = 15/2.
So the point of inflexion is (-1/2, 15/2).
This is how analysing the shape of the derivative graph can reveal how the original function bends.
Three layers of information
It helps to distinguish three related quantities.
f(x) describes the original output.
f'(x) describes how quickly that output changes with x.
f”(x) describes how the first derivative changes with x.
For a displacement-time model, these correspond to displacement, velocity and acceleration when the units and assumptions are suitable.
For a general algebraic curve, they describe the function value, gradient and changing gradient.
A learner should not assume a root of f” is automatically a turning point of f.
A root of f” may instead suggest a change in concavity, which must be checked.
Likewise, a root of f’ need not imply that f turns.
The different derivative levels answer different questions.
A derivative can touch zero without a turning point
Consider:
g(x) = (x – 1)³.
Differentiate:
g'(x) = 3(x – 1)².
The derivative equals zero at x = 1.
But the derivative is positive on both sides of x = 1.
Therefore g increases before and after the stationary point.
Its graph becomes horizontal at (1, 0) and then continues increasing.
This is a stationary point of inflexion.
The derivative graph is an upward-opening parabola touching the horizontal axis at x = 1 without passing below it.
The original function does not reverse direction.
This is why merely counting the zeros of a derivative graph is not enough to count turning points.
We must inspect whether its sign changes.
Compare crossing and touching in the derivative graph
For f'(x) = 6(x + 2)(x – 1), the derivative crosses the horizontal axis at -2 and 1.
Its sign changes at both crossings, so the original function has a turning point at each one.
For g'(x) = 3(x – 1)², the derivative touches the horizontal axis at 1 but stays non-negative.
There is no change from increasing to decreasing or vice versa in the original cubic.
Thus a derivative graph’s crossing behaviour matters.
A zero tells us that a gradient is zero.
A sign change tells us whether the direction of the original function changes.
This distinction prevents a common misconception: every stationary point is automatically a maximum or minimum.
What if the derivative is zero across an interval?
Suppose:
h'(x) = 0
for every x in an interval.
Then h is constant on that interval, assuming the usual differentiability conditions.
For example, if h(x) = 7, its derivative is zero everywhere.
But this is very different from a derivative that vanishes only at one isolated point.
For x³, the derivative is zero at the origin and positive elsewhere.
The function does not remain constant over any neighbourhood of zero.
A-Math students should learn to interpret the extent of a derivative condition.
“Zero at one input” and “zero throughout an interval” are different statements.
The corresponding conclusions about the original function differ.
A derivative graph may reveal a change without giving its amount
Suppose a derivative graph stays negative from x = -2 to x = 1.
That tells us the original function decreases throughout the interval.
But how much does it decrease?
If we know the precise derivative expression, we can integrate it.
For our cubic example:
f'(x) = 6x² + 6x – 12.
Calculate:
∫ from -2 to 1 of f'(x) dx.
An antiderivative is:
2x³ + 3x² – 12x.
At x = 1, this equals:
2 + 3 – 12 = -7.
At x = -2, it equals:
-16 + 12 + 24 = 20.
The difference is:
-7 – 20 = -27.
Therefore f(1) – f(-2) = -27.
This agrees with the original function values -6 and 21.
The signed area under the derivative graph tells us the net change of the original function.
Why the negative integral is sensible
Between -2 and 1, the derivative graph lies below the horizontal axis.
Therefore its definite integral over that interval is negative.
The original function decreases from 21 to -6.
Its change is:
-6 – 21 = -27.
The negative sign represents decreasing output.
It does not mean that the geometric area between the derivative curve and horizontal axis is a physically negative size.
The corresponding non-negative geometric area has magnitude 27 square units in pure coordinate interpretation.
If the derivative describes a physical rate, the integral’s unit depends on the output and input quantities.
The sign, integral and change in the original function all agree.
A graph’s derivative may not be defined everywhere
Consider:
f(x) = |x|.
For x < 0, the derivative is -1.
For x > 0, the derivative is 1.
At x = 0, the graph has a sharp corner and the usual derivative does not exist.
The original function has a minimum at zero.
But this minimum is not an ordinary differentiable stationary point where f'(0) = 0.
The derivative is undefined there.
This shows why graph analysis must also consider points where differentiability fails, not only zeros of a derivative.
In standard smooth polynomial questions, the derivative exists everywhere. For more general functions, the assumptions matter.
This is a useful enrichment connection to the limits of derivative-based classification.
Distinguish a discontinuity from an undefined derivative
A function may be continuous but not differentiable at a point, as with |x| at zero.
A function may also be undefined or discontinuous at a boundary, as with 1/x at zero.
These situations are not identical.
A corner can prevent a unique tangent gradient even when the graph passes continuously through the point.
An excluded denominator can prevent the function from having an ordinary value at all.
Students should therefore ask what the original function’s domain and graph allow before applying derivative tests.
Additional Mathematics builds this attention to conditions across algebra and calculus.
An invalid derivative calculation can arise from misunderstanding the function, not merely from forgetting a formula.
A derivative sign chart as a compact explanation
In a written solution, a sign chart can be very efficient.
For:
f'(x) = 6(x + 2)(x – 1),
record the critical inputs -2 and 1.
Use positive signs outside them and a negative sign between them.
Then write:
f increases on (-∞, -2), decreases on (-2, 1), and increases on (1, ∞).
At x = -2 there is a local maximum.
At x = 1 there is a local minimum.
This communicates the reasoning without repeating a full paragraph for every interval.
The sign chart is not decoration.
It makes the classification inspectable and helps the student notice if a candidate was omitted.
Read a derivative graph without its equation
Suppose a question provides only the graph of f'(x).
Imagine it crosses the horizontal axis at x = 0 from negative to positive, and later touches the axis at x = 3 without changing sign.
What can we conclude?
At x = 0, f changes from decreasing to increasing, so it has a local minimum, assuming it is differentiable around that point.
At x = 3, f'(3) = 0, but the graph remains positive on both sides.
So the original function is stationary there without a local turn.
If the derivative is otherwise positive for x > 0, f keeps increasing through x = 3.
But we cannot identify the original function’s absolute height at these points unless more information is provided.
This is a useful exercise in reading only what a graph supports.
A common error: treating the derivative graph as the original graph
A student sees the graph of f'(x) with a minimum at x = 2 and writes, “The original function has a minimum at x = 2.”
That conclusion is not justified.
The derivative’s minimum means f’ is smallest there.
It does not directly mean f is smallest.
A local minimum of f normally depends on f’ changing from negative to positive.
A minimum of f’ may instead correspond to a change in concavity of f.
To make a correct conclusion, ask whether the derivative value is positive, negative or zero at the point and how its sign behaves nearby.
The labels f and f’ cannot be treated as interchangeable.
Why a derivative may be negative but increasing
Consider:
f'(x) = x – 3.
For x = 1, the derivative is -2.
For x = 2, it is -1.
For x = 2.5, it is -0.5.
The derivative is increasing over these inputs but remains negative.
Therefore the original function is still decreasing, although its rate of decrease is becoming smaller in magnitude.
This is an important distinction.
“Derivative increasing” is not the same as “original function increasing”.
The derivative’s sign determines whether f increases or decreases.
The derivative’s own trend describes how the gradient changes.
That is why understanding second derivatives and concavity helps with full graph interpretation.
A function can increase while its derivative decreases
Take:
f(x) = √x,
for x > 0.
Its derivative is:
f'(x) = 1/(2√x).
This derivative is positive for all positive x, so f increases.
But f'(x) becomes smaller as x grows, so the graph gradually flattens.
For example:
At x = 1, f'(x) = 1/2.
At x = 4, f'(x) = 1/4.
At x = 9, f'(x) = 1/6.
The original output continues to rise while the rate of increase declines.
This is a useful idea when interpreting any model where progress continues but slows.
The relationship should not be applied to a real process without checking whether the mathematical model is appropriate.
A model’s derivative can help compare two quantities
Suppose two functions f and g describe comparable quantities with the same input variable.
The difference between their derivatives is:
f'(x) – g'(x).
That is the derivative of their difference:
d/dx[f(x) – g(x)] = f'(x) – g'(x).
If the two derivatives are equal at x = a, the difference function is stationary there.
But that fact alone does not tell us whether f(a) = g(a).
The original quantities can have different values while changing at the same rate.
This connects derivative-graph reading with multi-model comparisons.
It also reinforces the distinction between equality of values and equality of slopes.
Why a derivative graph is useful for checking a sketch
After sketching an original function, students can compare it with the derivative information.
Does the original curve rise where the derivative is positive?
Does it fall where the derivative is negative?
Are local turning points at locations where the derivative changes sign?
Does a flat crossing correspond to a derivative zero with no sign change?
Are curvature changes consistent with the derivative increasing or decreasing?
These checks can expose a sketching mistake even when the original formula is complicated.
A graph and its derivative should tell one coherent mathematical story.
Independent representations are useful precisely because they can confirm or challenge one another.
A five-session learning plan
Session one: connect x² and 2x
Sketch f(x) = x² and f'(x) = 2x.
Mark intervals where the derivative is positive and negative.
Describe how they match the parabola’s movement.
Session two: use a factorised derivative
Start with f'(x) = 6(x + 2)(x – 1).
Create a sign chart and predict local maxima and minima of f.
Then use the given cubic to calculate the coordinates.
Session three: investigate a stationary inflexion
Compare g(x) = (x – 1)³ with its derivative 3(x – 1)².
Explain why a derivative touching zero need not make the original function turn.
Session four: read signed accumulation
Integrate f'(x) from -2 to 1 and obtain -27.
Compare this with the difference between f(1) and f(-2).
Explain the negative sign.
Session five: interpret unfamiliar graphs
Use a supplied derivative graph without an equation.
Describe increasing and decreasing intervals, stationary candidates and what cannot be determined without an initial function value.
This progression develops interpretation, not just symbol manipulation.
What parents can ask without teaching differentiation
Ask the child what f'(x) measures.
Ask whether a derivative below zero means the original function’s output must be negative.
Ask why a zero derivative sometimes gives a turning point and sometimes does not.
Ask whether a derivative graph reveals the original function’s vertical position.
Ask how the area under a derivative graph relates to change in the original function.
These questions encourage the student to explain the relationship rather than memorise a single procedure.
A learner who can read a derivative sign chart independently has developed useful conceptual control.
Singapore’s Additional Mathematics context
The 2027 Singapore-Cambridge Secondary Education Certificate G3 Additional Mathematics syllabus, K341, includes differentiation, gradients, increasing and decreasing functions, stationary turning points and points of inflexion.
The derivative-graph reconstructions and some second-derivative interpretations in this article are teaching connections; students should follow the precise assessed methods in the official syllabus rather than assume every extension is separately examined.
For the current subject listing, use the SEAB 2027 G3 syllabus directory.
The worked functions are illustrative mathematical examples, not predictions of future examination questions.
How a three-student tutorial can help
One learner may differentiate correctly but confuse the derivative graph with the original function.
Another may make an accurate sign chart yet fail to classify a stationary inflexion.
A third may understand local behaviour but assume the derivative fixes the original function’s vertical position.
These students need different explanations.
In a three-student Additional Mathematics tutorial, a tutor can inspect the calculation, ask the student to interpret each sign and then test transfer using a changed function.
The immutable Secondary 1 Mathematics small-group tutorial in Clementi illustrates why early fluency with graphs, algebra and clear working matters before advanced calculus.
Students who are progressing well independently may not require tuition. The purpose of extra support is to repair a diagnosed learning obstacle.
Frequently asked questions
Does a positive derivative mean the function value is positive?
No. A positive derivative describes increasing behaviour; the function’s actual value may be positive, zero or negative.
Can a derivative graph reveal maxima and minima?
Yes, by showing where the derivative changes sign. A positive-to-negative change gives a local maximum, and negative-to-positive gives a local minimum.
Does every zero of the derivative give a turning point?
No. A derivative may touch zero without changing sign, producing a stationary point with no local reversal.
Can I recover the original function from its derivative graph?
You can recover a family of antiderivatives when the derivative function is known, but a constant or initial value is generally needed to determine a specific vertical position.
What does a negative definite integral of f’ mean?
It means the original function’s net change over the interval is negative. It need not describe a negative geometric area.
Is a minimum point on f’ also a minimum point on f?
Not necessarily. It describes the smallest gradient of f and may relate to a point of inflexion rather than an extremum of f.
Is the derivative graph useful outside examination questions?
Yes. It provides a way to reason about changing quantities and the speed and direction of change, provided the model and axis definitions are appropriate.
The lasting value of reading the second graph
Additional Mathematics gives students a way to see a function from more than one perspective.
The original graph displays values.
The derivative graph displays rates.
The second derivative describes how those rates change.
A learner who can connect them can predict increasing and decreasing behaviour, classify stationary points, interpret signed accumulation and recognise which information remains unknown.
This is more powerful than simply calculating one gradient.
It is a coherent way to understand change.
For connected reading, explore Why Study Additional Mathematics | Recognise Stationary Points That Are Not Turning Points, Why Study Additional Mathematics | Connect Secant Slopes to Instantaneous Rates, and the eduKateSG Additional Mathematics project library.