Calculus
Cambridge O Level Additional Mathematics 4037 Topic 14 revision chapter covering the whole of Calculus for the 2025-2027 examination cycle, teaching all fifteen official outcomes 14.1 to 14.15 in three connected parts: differentiation, integration and rectilinear kinematics. Part A opens by establishing the derived function as a gradient function rather than a symbolic rule, showing the secant gradient delta y over delta x approaching the tangent gradient as delta x tends to zero, and stating plainly that only an informal understanding of a limit is required and that differentiation from first principles is not examinable. Calculus notation is then fixed so that finite change and instantaneous rate are never confused: f dashed x, dy by dx, f double dashed x, d squared y by dx squared, d by dx of dy by dx, and the evaluation bar notation for a derivative at a stated value. The standard derivative bank for x to the n with rational n, sin x, cos x, tan x, e to the x, ln x and a constant is built with its domain conditions attached and with the radian requirement stated as a condition of validity rather than as a convention. The chain rule is taught as a layered structure, outer function then inner function then multiply by the inner derivative, worked through y equals the fourth power of three x squared plus one to give twenty four x times the cube of three x squared plus one, and extended to the linear-inside forms for sine, cosine, exponential and logarithm. Product and quotient rules follow with the required worked cases y equals x squared e to the x and y equals x plus one over x minus one, the latter carrying its restriction x not equal to one through the cancellation. Gradients, tangents and normals are then constructed from the point, the gradient and the negative reciprocal, including the horizontal-tangent and vertical-normal boundary case. Stationary points are located and given complete coordinates, and are classified by both the first-derivative sign test and the second-derivative test, with the inconclusive case f double dashed a equal to zero handled honestly and points of inflexion excluded as required content. Connected rates of change, small increments and linear approximation are taught with units and signs preserved, and practical optimisation is given an eight-step workflow that insists on a stated interval, a classification and an endpoint check rather than assuming every stationary point is a maximum. Part B builds integration as reverse differentiation, insisting on the arbitrary constant, covering the power rule, the logarithmic forms for one over x and one over ax plus b, the full standard composite bank for a linear inside, definite evaluation as F of b minus F of a, and the crucial distinction between signed accumulation and geometric area, with a worked area between the line y equals two x and the curve y equals x squared giving four thirds of a square unit. Part C connects position, velocity and acceleration by differentiation in one direction and integration with initial conditions in the other, separates displacement from distance by splitting at every turning time, and closes with a gradient-and-area dictionary for displacement-time, distance-time, velocity-time, speed-time and acceleration-time graphs, illustrated by a single cubic displacement model drawn consistently across all four graph types on one shared time axis. Fourteen original inline diagrams, thirty-six fully worked examples with boundary-case analyses, six comparison tables, a thirty-four-point mistake clinic, a retrieval check with accessible answer reveals, an exam-style mixed challenge, a mastery checklist and a spaced-review plan complete the chapter. All content is original and independent; the current Cambridge syllabus remains the authority for scope and assessment.Show moreShow less
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What is Calculus about?
The curve \(C\) has equation \(y=2x^3-9x^2+12x-3\).
A manufacturer makes an open cylindrical can of volume \(54\pi\ \mathrm{cm^3}\). The base has radius \(r\ \mathrm{cm}\) and the can has height \(h\ \mathrm{cm}\). The material used is proportional to the total external surface area \(S\ \mathrm{cm^2}\), which for an open can is \(S=\pi r^2+2\pi rh\).
A particle \(P\) moves in a straight line so that, \(t\) seconds after leaving the origin \(O\), its velocity is \(v=6t^2-30t+36\ \mathrm{m\,s^{-1}}\).
A particle \(Q\) starts from rest at a point \(A\) and moves in a straight line with acceleration \(a=\bigl(4-2t\bigr)\ \mathrm{m\,s^{-2}}\), where \(t\) is the time in seconds after it starts.
The twelve questions Differentiate \(f(x)=(2x-1)^5+\ln x\). Differentiate \(y=x^2\sin x\). Differentiate \(y=\dfrac{e^x}{x+1}\). Find the tangent and the normal to \(y=x^3\) at \(x=1\). Find and classify the stationary points of \(y=x^3-6x^2+9x+1\). Use small increments to estimate \(\sqrt[3]{8.12}\). A sphere has \(V=\tfrac43\pi r^3\). Find \(\dfrac{dV}{dt}\) when \(r=3\ \mathrm{cm}\) and \(\dfrac{dr}{dt}=0.2\ \mathrm{cm\,s^{-1}}\). Find \(\displaystyle\int\left(6x^2-\frac4x+3e^{2x}\right)dx\). Find \(\displaystyle\int\bigl\{\sin(3x)-2\sec^2(2x)\bigr\}dx\). Find the area of the region between \(y=2x\) and \(y=x^2\). A particle has \(v=3t^2-12t+9\) and \(s(0)=2\). Find its displacement and the distance travelled from \(t=0\) to \(t=4\). State what the gradient and the signed area represent on a velocity–time graph.
Key ideas to remember
- Every one of these fifteen outcomes reduces to the same question: am I being asked for the rate, or for the accumulation? Differentiate for the rate. Integrate for the accumulation. Everything else is bookkeeping.
- Only the last of the six is a failure of validity — degrees used where the rule needs radians. The other five are failures of finishing rather than of technique: the missing chain factor, the missing classification, the excluded case \(n=-1\), the question misread, and the interval left unsplit. Build the finishing habits and most of the topic looks after itself.
- Every error in drill 3 is one of exactly two kinds: a missing chain factor, or a rule applied to the wrong structure. If you diagnose those two reliably, you have most of Part A.
- Notice what Question A part (e) and Question B part (c) have in common: in each, the routine method produces something that does not fit, and the mark is for saying so clearly rather than for forcing an answer.
- Two sign facts to fix now, because they are mirror images and are constantly swapped. Differentiating \(\cos\) introduces a minus sign. Integrating \(\sin\) introduces a minus sign. Cosine loses the sign going one way; sine gains it going the other.
- Every finished integral in this lab can be checked in one line by differentiating it. Every finished area can be sanity-checked by asking whether the number is plausible for the size of the region on a sketch. Two habits, almost all of Part B's marks.
- Question A(d) and Question B(d) are the same idea from opposite sides: does the velocity change sign on this interval? If yes, split. If no, say so and take one integral. Answering that question first turns every distance part into routine arithmetic.
- Three of these eight questions have an answer of the form “there is none” or “the obvious method does not apply”. That proportion is not accidental: recognising when a routine does not fit is as examinable as executing one that does.
What you need to be able to do
- 14.1 — I can explain a derivative as a gradient function and as an instantaneous rate of change, and I understand informally what happens as \(\delta x\to0\).
- 14.2 — I use \(f'(x)\), \(\dfrac{dy}{dx}\), \(f''(x)\), \(\dfrac{d^2y}{dx^2}\) and \(\dfrac{d}{dx}\!\left(\dfrac{dy}{dx}\right)\) correctly, and I never confuse the finite change \(\delta y/\delta x\) with the instantaneous \(dy/dx\).
- 14.3 — I can differentiate \(x^n\) for rational \(n\), \(\sin x\), \(\cos x\), \(\tan x\), \(e^x\) and \(\ln x\), together with constants, constant multiples, sums and composite functions using the chain rule.
- 14.4 — I can differentiate products and quotients, and I can tell which rule a given expression actually needs.
- 14.5 — I can find the gradient at a point and write down the equation of the tangent and of the normal there.
- 14.6 — I can locate stationary points and state their complete coordinates.
- 14.7 — I can solve connected-rate problems, use small increments, and produce linear approximations with correct signs and units.
- 14.8 — I can set up and solve a practical maximum or minimum problem, including its valid interval and its endpoints.
- 14.9 — I can classify a stationary point as a maximum or a minimum using either the first- or the second-derivative test, and I can write the justification out in full.
- 14.10 — I understand integration as reverse differentiation and I include the arbitrary constant every time.
- 14.11 — I can integrate sums of terms involving powers of \(x\), including \(\dfrac1x\) and \(\dfrac1{ax+b}\).
- 14.12 — I can integrate \((ax+b)^n\), \(\sin(ax+b)\), \(\cos(ax+b)\), \(\sec^2(ax+b)\) and \(e^{ax+b}\), including the case \(n=-1\).
- 14.13 — I can evaluate a definite integral and use it to find a plane area, including the area between a line and a curve, between two curves, and as a sum of two regions.
- 14.14 — I can move between displacement, velocity and acceleration by differentiating or integrating, and I use initial conditions to fix every constant.
- 14.15 — I can draw and read displacement–time, distance–time, velocity–time, speed–time and acceleration–time graphs, and I know what gradient and signed area mean on each.
Key terms in Calculus
- Quotient Rule
- The rule for differentiating one function divided by another. If y equals u over v, where u and v are both functions of x and v is not zero, then dy by dx equals v times du by dx minus u times dv by dx, all divided by v squared. The order of the two terms in the numerator matters because subtraction is not commutative, and reversing them produces an answer with the wrong sign throughout. The restriction v not equal to zero comes from the original expression and must be carried into the answer. It is not supplied in the examination formula list.
- Derived Function
- The function that gives the gradient of a curve at every point of its domain, written f dashed of x or dy by dx. Its value at a particular x is the gradient of the tangent to the curve at that point, and equivalently the instantaneous rate at which y is changing with respect to x there. It is obtained as the limiting value of the gradient of a chord as the horizontal separation of the chord's two endpoints shrinks towards zero. For Cambridge O Level Additional Mathematics 4037 only an informal understanding of this limiting process is required; differentiation from first principles is not examinable.
- Second Derivative
- The derivative of the first derivative, written f double dashed of x or d squared y by dx squared. It measures the rate at which the gradient itself is changing, so it describes how a curve bends: a positive second derivative means the gradient is increasing and the curve is concave upwards, and a negative second derivative means the gradient is decreasing and the curve is concave downwards. Its main use in Cambridge O Level Additional Mathematics 4037 is to classify a stationary point as a maximum or a minimum, and in kinematics it gives acceleration as the second derivative of displacement with respect to time.
- Chain Rule
- The rule for differentiating a composite function, that is, a function of a function. If y equals f of g of x, then dy by dx equals f dashed of g of x multiplied by g dashed of x. In words: differentiate the outer function while leaving the inner function untouched, then multiply by the derivative of the inner function. The extra factor is needed because a change in x first changes the inner function and that change then changes y, so the two rates multiply. In Leibniz notation the same rule reads dy by dx equals dy by du multiplied by du by dx. It is not supplied in the examination formula list.
- Normal to a Curve
- The straight line through a point on a curve that is perpendicular to the tangent at that point. If the tangent has non-zero gradient m, the normal has gradient equal to minus one over m, because the product of the gradients of two perpendicular non-vertical lines is minus one. Where the tangent is horizontal the normal is vertical and has the equation x equals a rather than a gradient; where the tangent is vertical the normal is horizontal. The normal is found from the same derivative as the tangent, so no new differentiation is required once the gradient at the point is known.
- Stationary Point
- A point on a curve at which the gradient is zero, so that the tangent there is horizontal. It is found by solving the equation f dashed of x equals zero for x and then substituting each solution back into the original function to obtain the corresponding y coordinate. A stationary point is a point and must be stated as a complete pair of coordinates; an x value on its own does not answer the question. Solving f dashed of x equals zero locates stationary points but does not classify them, because a local maximum, a local minimum and a stationary point of inflexion all have a horizontal tangent.
- Second-Derivative Test
- A method of deciding whether a stationary point is a local maximum or a local minimum by evaluating the second derivative there. At a stationary value x equals a, if f double dashed of a is negative the curve is bending downwards and the point is a local maximum, and if f double dashed of a is positive the curve is bending upwards and the point is a local minimum. If f double dashed of a is zero the test gives no information and a different method must be used, such as examining the sign of the first derivative on either side of the stationary value. A complete justification states the stationary value, shows the second derivative evaluated at it, and draws the conclusion.
- Optimisation
- The use of differentiation to find the greatest or least value a quantity can take in a practical situation. The quantity to be optimised is first expressed as a function of a single variable, using a constraint from the problem to eliminate any other variables, and the interval of values that variable may legitimately take is stated. The derivative is then set to zero to find candidate values, each candidate is classified or compared, and the endpoints of the interval are examined because the greatest value of a function on a closed interval can occur at an endpoint rather than at a stationary point. The final answer is the quantity the question asked for, given with its units.
- Composite Linear Integral
- An integral in which the integrand is a standard function applied to a linear expression a x plus b. Because the derivative of a linear inside is the constant a, reversing the chain rule requires only division by that constant, so the antiderivative is the ordinary antiderivative of the standard function evaluated at a x plus b and then divided by a. This gives the standard forms for the power, sine, cosine, secant squared and exponential functions of a linear argument, each with a factor one over a and an arbitrary constant. The method is valid only when the inside is linear; a non-linear inside has a non-constant derivative that cannot be divided out in this way.
- Indefinite Integral
- The family of all functions whose derivative is a given function. If F dashed of x equals f of x, then the indefinite integral of f with respect to x is written as F of x plus C, where C is an arbitrary constant. The constant is required because differentiating any constant gives zero, so infinitely many functions differing only by a constant share the same derivative, and reversing the process cannot recover which one was intended. An indefinite integral is a family of functions, not a number, and any indefinite integral can be checked by differentiating the result to see whether the original function is recovered.
- Connected Rates of Change
- A technique for finding an unknown rate of change from a known one when two quantities are linked by a formula. If A depends on r and r depends on time, then the rate at which A changes with time is the product of the rate at which A changes with r and the rate at which r changes with time, written dA by dt equals dA by dr multiplied by dr by dt. The method is the chain rule applied with time as the underlying variable. Every rate carries units built from the two quantities involved, and a negative rate means the quantity is decreasing, so both units and signs form part of a complete answer.
- Rectilinear Motion
- Motion of a particle along a straight line, described by three signed quantities that are linked by calculus. Displacement is the particle's position relative to a fixed origin, velocity is the rate of change of displacement with respect to time, and acceleration is the rate of change of velocity with respect to time and therefore the second derivative of displacement. Reversing the chain, velocity is the integral of acceleration and displacement is the integral of velocity, with each constant of integration determined by an initial condition. Distance travelled and speed are the unsigned counterparts of displacement and velocity, and they differ from them whenever the particle changes direction.
- Velocity-Time Graph
- A graph of a particle's velocity against time, from which two further quantities can be read directly. Its gradient at any instant is the acceleration at that instant, because acceleration is the rate of change of velocity with respect to time. The signed area between the graph and the time axis over an interval is the displacement over that interval, with area below the axis counting negatively because the particle is then moving in the negative direction. Distance travelled is obtained instead from the corresponding speed-time graph, in which the parts of the velocity graph below the axis are reflected above it, so that all area counts positively.
- Definite Integral
- The number obtained by evaluating an antiderivative at the upper limit and subtracting its value at the lower limit, written as the integral from a to b of f of x with respect to x, and equal to F of b minus F of a. No arbitrary constant appears, because any constant added to the antiderivative cancels in the subtraction. A definite integral measures signed accumulation: where the curve lies above the horizontal axis the contribution is positive and where it lies below the axis the contribution is negative. It therefore equals the geometric area of the region between the curve and the axis only when the curve does not change sign across the interval of integration.
- Logarithmic Integral
- The antiderivative of one over x, equal to the natural logarithm of the modulus of x plus an arbitrary constant. It is the single case the power rule for integration cannot handle, because raising the index minus one by one gives zero and the rule would require division by zero. The modulus signs are needed so that the antiderivative is valid on intervals where x is negative as well as where it is positive, since the natural logarithm itself is only defined for positive arguments. The related form one over a x plus b integrates to one over a times the natural logarithm of the modulus of a x plus b, plus a constant, provided a is not zero.
Common mistakes to avoid
- 1. “\(\dfrac{d}{dx}(3x^2+1)^4 = 4(3x^2+1)^3\)” Why it failsThe power rule differentiates a power of \(x\). Here the thing being raised to a power is itself a function of \(x\), so changing \(x\) changes the bracket, and that change has to be accounted for. CorrectMultiply by the derivative of the inside: \(4(3x^2+1)^3\times 6x=24x(3x^2+1)^3\). Detect itIf the bracket contains anything other than a bare \(x\), a chain-rule factor is owed.
- 2. “\(\dfrac{dy}{dx}=0\), so this is a maximum.” Why it fails\(f'(x)=0\) says the tangent is horizontal. A maximum, a minimum and a stationary point of inflexion all have horizontal tangents, so the equation cannot distinguish between them. CorrectEvaluate \(f''\) at the stationary value, or test the sign of \(f'\) either side, and state the conclusion that follows. Detect itAny sentence containing “maximum” that is not preceded by an evaluated test is unsupported.
- 3. “\(\displaystyle\int \frac1x\,dx = \frac{x^0}{0}+C\)” Why it failsThe power rule \(\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}+C\) requires \(n\ne-1\), for the visible reason that \(n+1=0\) makes the denominator zero. \(1/x\) is precisely the excluded case. Correct\(\displaystyle\int\frac1x\,dx=\ln|x|+C\), valid on any interval not containing \(0\). Detect itA zero denominator appearing from nowhere is always this error.
- 4. “The integral came out negative, so I made a sign error.” Why it failsA definite integral is a signed accumulation. Where the curve is below the axis it contributes negatively, entirely correctly. A negative value is information, not a mistake. CorrectIf a geometric area is wanted, find every crossing of the axis, integrate between consecutive crossings, and add the magnitudes. Detect itAsk whether the question said “evaluate the integral” or “find the area”. They are different requests.
- 5. “Displacement from \(t=0\) to \(t=4\) is \(4\), so it travelled \(4\).” Why it failsDisplacement is the signed change in position; distance is the total path length. If the particle reversed direction, some of the outward journey is cancelled by the return. CorrectSolve \(v=0\) for the turning times inside the interval, evaluate \(s\) at every turning time and at both ends, and add the absolute changes. Detect itCheck whether \(v\) changes sign on the interval. If it does, the two answers differ.
- 6. “\(\dfrac{d}{dx}\sin x=\cos x\), so \(\sin 30\) differentiates to \(\cos 30\).” Why it failsThe standard derivatives are proved using a limit that only holds when the angle is in radians. In degrees the derivative of \(\sin x\) is \(\frac{\pi}{180}\cos x\), so every subsequent value is wrong by that factor. CorrectConvert to radians before doing any trigonometric calculus, and put the calculator into radian mode for Paper 2. Detect itA degree symbol anywhere near a derivative or an integral is a warning sign.
How Calculus is examined
- Calculus questions are usually structured: several short parts that build on one another. That structure is an advantage — the early parts hand you the pieces the later parts need — but it also means an error in part (a) propagates. This section is about reading the question, not about the mathematics.
- The syllabus defines a fixed list of command words, and each one means the same thing wherever it appears. These are the ones that recur in calculus questions.
- Differentiate, find stationary points, classify them, then find a tangent or normal, then perhaps integrate for an area under the same curve. One curve, five parts.
- The trap: reusing \(f'(x)\) where \(f(x)\) is needed, having spent three parts working with the derivative.
- A box, a fence, a cylinder or a cost. Build the quantity as a function of one variable using a constraint, differentiate, solve, justify, answer with units.
- The trap: forgetting the constraint that eliminates the second variable, then differentiating a two-variable expression.
Frequently asked questions
What is a derivative?
A derivative, written \(f'(x)\) or \(\dfrac{dy}{dx}\), is the gradient function of a curve: its value at a point is the gradient of the tangent there, and equivalently the instantaneous rate at which \(y\) is changing with respect to \(x\). It comes from the limiting value of the gradient of a chord as the two endpoints move together, though only an informal understanding of that limit is required — differentiation from first principles is not examinable. \(f'(x)\) is a new function derived from \(f(x)\), not another name for it.
Why do you need the chain rule to differentiate \((3x^2+1)^4\)?
Because the power rule alone differentiates a power of \(x\), and here the quantity being raised to a power is itself a function of \(x\), a bracket that changes as \(x\) changes. The chain rule accounts for that inner change: differentiate the outer power to get \(4(3x^2+1)^3\), then multiply by the derivative of the inner function, \(6x\), giving \(24x(3x^2+1)^3\). Forgetting that extra factor is the single most common differentiation error in this chapter.
Why doesn't \(f'(x)=0\) alone tell you whether a stationary point is a maximum or a minimum?
Because a maximum, a minimum and a stationary point of inflexion all have a horizontal tangent, so \(f'(x)=0\) is satisfied by all three and cannot distinguish between them. Classify the point separately: evaluate \(f''(x)\) there, where a negative value gives a maximum and a positive value gives a minimum, or test the sign of \(f'(x)\) on either side if \(f''(x)=0\). A complete justification states the stationary value, the test used, and the conclusion.
Why can't you integrate \(\dfrac1x\) using the ordinary power rule?
Because the power rule \(\displaystyle\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C\) requires \(n\ne-1\), since \(n=-1\) makes the denominator \(n+1\) equal to zero. \(\dfrac1x\) is exactly \(x^{-1}\), the excluded case, so the rule cannot be applied to it directly. Instead \(\displaystyle\int\dfrac1x\,dx=\ln|x|+C\), and more generally \(\displaystyle\int\dfrac1{ax+b}\,dx=\dfrac1a\ln|ax+b|+C\); the modulus is needed because the logarithm itself is only defined for a positive argument.
Why is a negative value from a definite integral not automatically a sign error?
Because a definite integral is a signed accumulation: where the curve lies below the horizontal axis, its contribution to \(\displaystyle\int_a^b f(x)\,dx\) is genuinely negative, not a mistake. It only equals the geometric area of a region when the curve does not change sign across the interval. If a geometric area is wanted, find every point where the curve crosses the axis, integrate each piece separately, and add the moduli of the separate values together.
Why isn't the distance a particle travels the same as its displacement?
Displacement is the signed change in position between two times, found from \(\displaystyle\int v\,dt\); distance is the total length of path covered, regardless of direction. If a particle reverses direction during the interval, part of the outward journey is cancelled by the return when you compute displacement, so the two numbers differ. Solve \(v=0\) to find any turning times inside the interval, integrate each separate stage, and add the moduli of those separate displacements to get the distance.
Why does differentiating \(\sin x\) in degree mode give the wrong derivative?
Because the standard derivatives \(\dfrac{d}{dx}\sin x=\cos x\) and \(\dfrac{d}{dx}\cos x=-\sin x\) are proved using a limit that only holds when \(x\) is measured in radians. In degree mode the derivative of \(\sin x\) is actually \(\dfrac{\pi}{180}\cos x\), so every value calculated from the degree-mode derivative is wrong by that constant factor. Always work calculus involving trigonometric functions in radians, and convert only at the very end if a degree answer is required.
Syllabus reference and sources
Written against: Cambridge O Level Additional Mathematics (4037) 2025–2027 Syllabus (Subject Content, Topic 14: Calculus).
Written by: Academiq Edu Instructor Panel
Source documents
- Cambridge O Level Additional Mathematics 4037 syllabus for 2025, 2026 and 2027
- Syllabus update notice, Cambridge O Level Additional Mathematics 4037, 2025–2027
- Cambridge O Level Additional Mathematics 4037 syllabus for 2028, 2029 and 2030 (version 1), consulted only to confirm that no significant change affects this topic
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