Topic 5: Calculus | IB Maths AA

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🎯What you need to be able to do

Interpret the derivative both as the gradient function and as a rate of change.
Differentiate powers, and the standard functions \( \sin x, \cos x, \mathrm{e}^{x}, \ln x \).
Apply the chain, product and quotient rules, including in combination.
Find equations of tangents and normals at a point.
Use the first and second derivatives to locate and classify stationary points and points of inflexion.
Solve optimisation problems in context, including checking that the stationary point is the right kind.
Handle kinematics with calculus: displacement, velocity, acceleration, and distance versus displacement.
Integrate as the reverse of differentiation, with a boundary condition to fix the constant.
Integrate by inspection and by substitution, including composites with \( ax+b \).
Find areas under a curve and between curves, handling regions below the axis correctly.
AHL Differentiate from first principles; use l’Hôpital’s rule for indeterminate limits.
AHL Use implicit differentiation and related rates.
AHL Integrate by parts and by substitution, and use partial fractions on the integrand.
AHL Find volumes of revolution, solve first-order differential equations, and build Maclaurin series.

📚The mathematics

5.1–5.4 Differentiation

The derivative is the gradient of the tangent, and equivalently the instantaneous rate of change. Both readings matter: \( \dfrac{\mathrm{d}y}{\mathrm{d}x} \) is a slope on a graph and a rate in a context, and questions switch between the two without warning. Notation: \( \dfrac{\mathrm{d}y}{\mathrm{d}x} \), \( f'(x) \), and in context \( \dfrac{\mathrm{d}V}{\mathrm{d}r} \) or \( \dfrac{\mathrm{d}s}{\mathrm{d}t} \).

\[ f(x) = ax^{n} \;\Longrightarrow\; f'(x) = anx^{\,n-1} \]

Before differentiating anything awkward, rewrite it as a power: \( \dfrac{3}{x^{2}} = 3x^{-2} \) and \( \sqrt{x} = x^{1/2} \). Most “I cannot differentiate this” moments are really “I have not rewritten this yet” moments.

\( f'(x) > 0 \) means increasing, \( f'(x) < 0 \) decreasing, and \( f'(x) = 0 \) stationary. For the tangent at \( x = a \), the gradient is \( f'(a) \) and the point is \( (a, f(a)) \); the normal is perpendicular, so its gradient is \( -\dfrac{1}{f'(a)} \). Two things trip people here: substituting into \( f' \) rather than \( f \) to get the \(y\)-coordinate, and forgetting the negative reciprocal for the normal.

5.6 The three rules

\( \dfrac{\mathrm{d}}{\mathrm{d}x}\left[f(g(x))\right] = f'(g(x))\,g'(x) \)

\( (uv)' = u'v + uv' \)

\( \left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^{2}} \)

With the standard derivatives \( \dfrac{\mathrm{d}}{\mathrm{d}x}\sin x = \cos x \), \( \dfrac{\mathrm{d}}{\mathrm{d}x}\cos x = -\sin x \), \( \dfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{e}^{x} = \mathrm{e}^{x} \) and \( \dfrac{\mathrm{d}}{\mathrm{d}x}\ln x = \dfrac{1}{x} \).

Identify the structure before you start. Is the expression a function of a function (chain), a product of two functions (product), or one divided by another (quotient)? Composite expressions frequently need two rules nested, and the chain rule is the one people forget on the inside — the derivative of \( \sin(3x) \) is \( 3\cos(3x) \), not \( \cos(3x) \).

In the quotient rule the order of the numerator matters, because subtraction is not commutative. A quotient with a simple denominator is often faster rewritten as a product with a negative index.

5.7 & 5.8 Second derivatives, stationary points and optimisation

At a stationary point \( f'(x) = 0 \). To classify it, either test the sign change of \( f' \) either side, or use the second derivative: \( f''(x) > 0 \) means a minimum (the curve is concave up, holding water), \( f''(x) < 0 \) means a maximum.

A point of inflexion is where the concavity changes, so \( f''(x) = 0 \) and \( f'' \) changes sign. The second condition is essential: \( y = x^{4} \) has \( f''(0) = 0 \) but no inflexion there, since the curve is concave up on both sides.

Optimisation questions in context follow a fixed procedure. Write the quantity to be optimised; use the constraint to eliminate the second variable so you have a function of one variable; state a sensible domain; differentiate; set to zero; solve; classify; and then answer the question that was actually asked, which is often the maximum value rather than the \(x\) that produces it. The classification step is worth a mark and is the one most often skipped.

5.9 Kinematics

\( v = \dfrac{\mathrm{d}s}{\mathrm{d}t} \)

\( a = \dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}} \)

\( s = \displaystyle\int v\,\mathrm{d}t \)

Differentiate going down the list, integrate coming back up. Two distinctions carry most of the marks. Speed is \( |v| \), so a particle moving in the negative direction has negative velocity but positive speed; the particle is at rest when \( v = 0 \) and changes direction when \( v \) changes sign.

And displacement is not distance:

displacement \( = \displaystyle\int_{t_1}^{t_2} v(t)\,\mathrm{d}t \)

distance \( = \displaystyle\int_{t_1}^{t_2} |v(t)|\,\mathrm{d}t \)

If the velocity changes sign inside the interval, those two answers differ. For distance, find where \( v = 0 \), split the integral there, and add the magnitudes.

5.5, 5.10 & 5.11 Integration

\[ \int x^{n}\,\mathrm{d}x = \frac{x^{n+1}}{n+1} + C, \quad n \ne -1; \qquad \int \frac{1}{x}\,\mathrm{d}x = \ln|x| + C \]

Add one to the index and divide by the new index — except at \( n = -1 \), where that would divide by zero, and the logarithm steps in. Never omit \( +C \) on an indefinite integral; when a boundary condition is given, substitute it to find \( C \) and give the particular function.

For composites with a linear function, integrate as usual and divide by the coefficient of \(x\): \( \int \cos(2x+3)\,\mathrm{d}x = \tfrac{1}{2}\sin(2x+3) + C \). More generally, reverse the chain rule (integration by inspection) or substitute when the integrand has the shape \( \int k\,g'(x)f(g(x))\,\mathrm{d}x \) — that is, when the derivative of the inside is sitting there as a factor.

The definite integral \( \int_a^b f'(x)\,\mathrm{d}x = f(b) - f(a) \) gives the signed area, and this is where marks disappear:

Area below the axis integrates to a negative number. If you are asked for an area and the curve crosses the \(x\)-axis in the interval, integrating straight through makes the parts cancel. Find the roots, split the integral at each, and add the absolute values. For the area between two curves, integrate \( \int (\text{upper} - \text{lower}) \) over the interval between their intersections — and if they cross, split there too.

AHL 5.12 & 5.13 Limits and first principles

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

You are expected to use this definition for polynomials: expand, cancel the \(h\) in the denominator, then let \( h \to 0 \). The cancelling step is the whole argument, because before it the quotient is \( \tfrac{0}{0} \).

For limits of the indeterminate forms \( \tfrac{0}{0} \) and \( \tfrac{\infty}{\infty} \), l’Hôpital’s rule says differentiate the numerator and the denominator separately and try again, repeating as needed. It is not the quotient rule, and you must confirm the indeterminate form before applying it.

AHL 5.14 & 5.15 Implicit differentiation and further derivatives

When \(y\) is not isolated, differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\) so that every \(y\) term picks up a factor of \( \dfrac{\mathrm{d}y}{\mathrm{d}x} \) by the chain rule. Then collect and solve for the derivative.

Related rates link two changing quantities through the chain rule, \( \dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{\mathrm{d}V}{\mathrm{d}r} \times \dfrac{\mathrm{d}r}{\mathrm{d}t} \). Write down what you are given and what you want as derivatives-with-respect-to-time first; the equation to differentiate is then obvious.

Further derivatives to know: \( \tan x, \sec x, \csc x, \cot x, a^{x}, \log_a x \), and the inverse trigonometric functions, together with the corresponding integrals — including the two that produce \( \arctan \) and \( \arcsin \) forms.

AHL 5.16 & 5.17 Further integration

Integration by parts reverses the product rule:

\[ \int u\,\frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x = uv - \int v\,\frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x \]

Choose \(u\) to be the part that gets simpler when differentiated — a polynomial, or \( \ln x \), or an inverse trigonometric function. Products of an exponential and a trigonometric function need parts twice, after which the original integral reappears and you solve for it algebraically.

Volumes of revolution about the axes:

\( V = \displaystyle\int_a^b \pi y^{2}\,\mathrm{d}x \)

\( V = \displaystyle\int_a^b \pi x^{2}\,\mathrm{d}y \)

Note it is \( y^{2} \), not \( y \) — squaring first, then integrating.

AHL 5.18 Differential equations

Three techniques. Separation of variables when the equation factorises as \( \dfrac{\mathrm{d}y}{\mathrm{d}x} = f(x)g(y) \): gather the \(y\) terms one side, the \(x\) terms the other, and integrate both, remembering a single constant. Homogeneous equations of the form \( \dfrac{\mathrm{d}y}{\mathrm{d}x} = f\!\left(\dfrac{y}{x}\right) \) yield to the substitution \( y = vx \), which requires the product rule on \( \dfrac{\mathrm{d}y}{\mathrm{d}x} = v + x\dfrac{\mathrm{d}v}{\mathrm{d}x} \). And linear equations \( y' + P(x)y = Q(x) \) are solved with the integrating factor \( \mathrm{e}^{\int P(x)\,\mathrm{d}x} \).

Euler’s method approximates numerically: \( y_{n+1} = y_n + h\,f(x_n, y_n) \) with \( x_{n+1} = x_n + h \). It is a spreadsheet task, and its accuracy improves as \(h\) shrinks.

AHL 5.19 Maclaurin series

\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3} + \cdots \]

You should know the standard expansions of \( \mathrm{e}^{x}, \sin x, \cos x, \ln(1+x) \) and \( (1+x)^{p} \), and be able to reach others by substituting, multiplying, differentiating or integrating a known series — almost always faster than differentiating from scratch four times.

✏️Worked example

A closed cylindrical can has volume 500 cm\(^3\). Its radius is \(r\) cm and its height \(h\) cm. (a) Show that the total surface area is \( A = 2\pi r^{2} + \dfrac{1000}{r} \). (b) Find the value of \(r\) that minimises \(A\), and justify that it is a minimum.

(a) The volume constraint gives \( \pi r^{2}h = 500 \), so \( h = \dfrac{500}{\pi r^{2}} \). A closed cylinder has two circular ends and a curved surface: \( A = 2\pi r^{2} + 2\pi rh \). Substituting for \(h\):

\[ A = 2\pi r^{2} + 2\pi r \times \frac{500}{\pi r^{2}} = 2\pi r^{2} + \frac{1000}{r} \]

as required. Note the \( \pi \) and one power of \(r\) cancel — if they do not, check the constraint.

(b) Rewrite as a power before differentiating: \( A = 2\pi r^{2} + 1000r^{-1} \), so

\[ \frac{\mathrm{d}A}{\mathrm{d}r} = 4\pi r - \frac{1000}{r^{2}} \]

Setting this to zero: \( 4\pi r = \dfrac{1000}{r^{2}} \), so \( r^{3} = \dfrac{1000}{4\pi} = 79.58 \) and \( r = 4.30 \) cm.

To justify the minimum, use the second derivative: \( \dfrac{\mathrm{d}^{2}A}{\mathrm{d}r^{2}} = 4\pi + \dfrac{2000}{r^{3}} \). For any \( r > 0 \) both terms are positive, so \( A'' > 0 \) and the stationary point is a minimum.

Check it. At \( r = 4.30 \), \( h = \dfrac{500}{\pi(4.30)^{2}} = 8.60 \) cm, which is exactly \( 2r \). The optimal closed cylinder always has height equal to diameter — it is as close to a sphere as a cylinder can get. If your \(h\) does not come out at \( 2r \), you have made an algebraic slip.

“Justify” is a separate mark from “find”. Solving \( A' = 0 \) locates a stationary point; it does not tell you which kind. Either show \( A'' > 0 \), or test the sign of \( A' \) either side. And in part (a), “show that” means you must derive the printed expression — substituting the given formula back into the volume to check it is circular reasoning and earns nothing.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Find the equation of the normal to \( y = x^{3} - 4x + 1 \) at the point where \( x = 2 \).

The point: \( y = 8 - 8 + 1 = 1 \), so \( (2, 1) \). The gradient: \( \dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2} - 4 \), which at \( x = 2 \) is \( 12 - 4 = 8 \). The normal gradient is \( -\tfrac{1}{8} \). So \( y - 1 = -\tfrac{1}{8}(x - 2) \), or \( x + 8y - 10 = 0 \). The two classic slips are substituting into the derivative to get the \(y\)-coordinate, and using 8 rather than \( -\tfrac{1}{8} \).

2. Differentiate \( y = x^{2}\mathrm{e}^{3x} \).

A product, and the second factor needs the chain rule too. With \( u = x^{2} \) and \( v = \mathrm{e}^{3x} \): \( u' = 2x \) and \( v' = 3\mathrm{e}^{3x} \). So \( \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x\mathrm{e}^{3x} + x^{2}(3\mathrm{e}^{3x}) = x\mathrm{e}^{3x}(2 + 3x) \). Factorising at the end is worth doing: it makes the stationary points visible immediately, at \( x = 0 \) and \( x = -\tfrac{2}{3} \).

3. A particle moves with velocity \( v(t) = t^{2} - 5t + 6 \) m s\(^{-1}\) for \( 0 \le t \le 4 \). Find the displacement and the total distance travelled.

Displacement is \( \int_0^4 (t^{2} - 5t + 6)\,\mathrm{d}t = \left[\tfrac{t^{3}}{3} - \tfrac{5t^{2}}{2} + 6t\right]_0^4 = \tfrac{64}{3} - 40 + 24 = \tfrac{16}{3} = 5.33 \) m.
For distance, first find where \( v = 0 \): \( (t-2)(t-3) = 0 \), so \( t = 2 \) and \( t = 3 \), both inside the interval. Splitting: \( \int_0^2 v = \tfrac{8}{3} - 10 + 12 = \tfrac{14}{3} \); \( \int_2^3 v = -\tfrac{1}{6} \); \( \int_3^4 v = \tfrac{5}{6} \). Total distance \( = \tfrac{14}{3} + \tfrac{1}{6} + \tfrac{5}{6} = \tfrac{17}{3} = 5.67 \) m. The two answers differ precisely because the particle reversed between \(t = 2\) and \(t = 3\).

4. Find the area enclosed between \( y = x^{2} \) and \( y = 2x + 3 \).

Find the intersections: \( x^{2} = 2x + 3 \) gives \( x^{2} - 2x - 3 = 0 \), so \( (x-3)(x+1) = 0 \) and \( x = -1, 3 \). Between these the line is above the parabola (test \( x = 0 \): line 3, curve 0). So area \( = \int_{-1}^{3}\left(2x + 3 - x^{2}\right)\mathrm{d}x = \left[x^{2} + 3x - \tfrac{x^{3}}{3}\right]_{-1}^{3} = (9 + 9 - 9) - (1 - 3 + \tfrac{1}{3}) = 9 - \left(-\tfrac{5}{3}\right) = \tfrac{32}{3} \). Because you integrated upper minus lower, the answer is positive automatically — no absolute values needed.

5. AHL Find \( \displaystyle\int x\ln x\,\mathrm{d}x \).

Integration by parts. Choose \( u = \ln x \) (it simplifies on differentiating; \(x\) would not) and \( \dfrac{\mathrm{d}v}{\mathrm{d}x} = x \). Then \( \dfrac{\mathrm{d}u}{\mathrm{d}x} = \dfrac{1}{x} \) and \( v = \dfrac{x^{2}}{2} \). So \( \int x\ln x\,\mathrm{d}x = \dfrac{x^{2}}{2}\ln x - \int \dfrac{x^{2}}{2}\cdot\dfrac{1}{x}\,\mathrm{d}x = \dfrac{x^{2}}{2}\ln x - \int \dfrac{x}{2}\,\mathrm{d}x = \dfrac{x^{2}}{2}\ln x - \dfrac{x^{2}}{4} + C \). Choosing \( u = x \) instead leaves you with \( \int \tfrac{x^2}{2}\ln x \), which is worse than where you started.

6. AHL Solve \( \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2xy}{x^{2}+1} \) given that \( y = 4 \) when \( x = 0 \).

Separable. Divide by \(y\) and multiply by \( \mathrm{d}x \): \( \displaystyle\int \dfrac{1}{y}\,\mathrm{d}y = \int \dfrac{2x}{x^{2}+1}\,\mathrm{d}x \). The right-hand integrand has the derivative of the denominator on top, so it integrates to a logarithm: \( \ln|y| = \ln\left(x^{2}+1\right) + c \). Exponentiating, \( y = A\left(x^{2}+1\right) \) where \( A = \mathrm{e}^{c} \). Applying the condition \( y(0) = 4 \): \( 4 = A(1) \), so \( A = 4 \) and \( y = 4\left(x^{2}+1\right) \). Combining the constants into a single multiplicative \(A\) before substituting is much cleaner than carrying \( \mathrm{e}^{c} \) through.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

3Blue1Brown — Essence of Calculus, the best available intuition for what a derivative is
Paul’s Online Math Notes — worked integration technique examples in quantity
Desmos — graph \( f \), \( f' \) and \( f'' \) together and watch the stationary points line up with the zeros