HookThe gradient that measured the fastest human alive
On 16 August 2009 in Berlin, Usain Bolt ran 100 m in 9.58 seconds — still the world record. What is less famous is that the German athletics federation had lined the track with laser and high-speed camera systems that logged his position roughly every hundredth of a second. Bolt's speed was never measured directly; there is no sensor for it. Instead the biomechanics team took his position-time data and asked, at each instant, how fast is the position changing right now? — the gradient of the position-time graph at that point. That gradient peaked at about \(12.3\) m/s (roughly \(44\) km/h) somewhere between the 60 and 80 metre marks, and it is the single most-quoted number from the race. It is also, exactly, a derivative.
That is the whole idea of Chapter 7. Differentiation turns a function that tells you where something is into a function that tells you how fast it is changing — and then does it again to tell you whether that change is speeding up or slowing down. You will build the derivative from its definition as a limit of gradients (first principles), learn the standard results that let you skip that limit for \(x^n\), \(e^{kx}\), \(\sin kx\), \(\cos kx\), \(\tan kx\) and \(\ln x\), use it to find tangents, normals, maxima and minima, unlock the product, quotient and chain rules for combinations, differentiate curves given implicitly and parametrically, and finally run the process in reverse-of-modelling: writing down a differential equation from a sentence like 'the population grows at a rate proportional to its size'. Every one of these is Bolt's gradient in a different costume.
ModelThe derivative from first principles — why the gradient is a limit
The gradient of a straight line is a single number. The gradient of a curve is different at every point, so we define it at a point \(P\) as the gradient of the tangent there. To pin that down, take a second point \(Q\) a small horizontal distance \(h\) away and compute the gradient of the chord \(PQ\): \(\dfrac{f(x+h)-f(x)}{h}\). As \(h\) shrinks towards zero, \(Q\) slides into \(P\) and the chord's gradient closes in on the tangent's. That limiting value is the derivative, written \(f'(x)\) or \(\dfrac{dy}{dx}\):
\[f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.\]
The derivative has two readings that examiners use interchangeably: it is the gradient of the tangent to \(y=f(x)\), and it is the rate of change of \(y\) with respect to \(x\) — which is why Bolt's velocity is \(\dfrac{ds}{dt}\), the rate of change of displacement with time. Differentiate a second time and you get \(f''(x)\), the rate of change of the gradient: where \(f''(x)>0\) the curve bends upward (convex), where \(f''(x)<0\) it bends downward (concave), and a point of inflection is where \(f''\) changes sign — where the bending flips. Sketching the gradient function \(y=f'(x)\) under \(y=f(x)\) is then just reading off where the original is rising (\(f'>0\)), falling (\(f'<0\)) or flat (\(f'=0\)).
Differentiate \(f(x)=x^2\) from first principles.
Start from the definition: \(f'(x)=\lim_{h\to 0}\dfrac{(x+h)^2-x^2}{h}\). Expand the top: \((x+h)^2-x^2=x^2+2xh+h^2-x^2=2xh+h^2\). So the chord gradient is \(\dfrac{2xh+h^2}{h}\). Now — and this is the one legal move — cancel the \(h\) before letting it vanish, because while \(h\neq 0\) you may divide by it: \(\dfrac{h(2x+h)}{h}=2x+h\). Only now take the limit: as \(h\to 0\), \(2x+h\to 2x\). Hence \(f'(x)=2x\). The same expansion on \(x^3\) gives \(3x^2\), and the pattern \(\dfrac{d}{dx}(x^n)=nx^{n-1}\) is what the next block lets you quote without redoing this limit each time. Note the trap: you cannot set \(h=0\) at the chord stage, because \(\dfrac{0}{0}\) is meaningless — you must simplify to \(2x+h\) first, then take the limit.
MechanismThe standard derivatives — the results you differentiate on sight
First principles proves the rules; in the exam you quote them. The core list you must know cold: \(\dfrac{d}{dx}(x^n)=nx^{n-1}\) for any rational \(n\) (so it covers \(\sqrt{x}=x^{1/2}\) and \(\dfrac{1}{x^2}=x^{-2}\)); \(\dfrac{d}{dx}(e^{kx})=ke^{kx}\); \(\dfrac{d}{dx}(a^{kx})=ka^{kx}\ln a\); \(\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}\); and for trigonometry in radians, \(\dfrac{d}{dx}(\sin kx)=k\cos kx\), \(\dfrac{d}{dx}(\cos kx)=-k\sin kx\) and \(\dfrac{d}{dx}(\tan kx)=k\sec^2 kx\).
Two linearity rules do the rest: you may differentiate a sum or difference term by term, and a constant multiple pulls straight through, so \(\dfrac{d}{dx}\big(5\sin 2x\big)=5\cdot 2\cos 2x=10\cos 2x\). The single most common wreck here is the trig one: these derivatives are only true when \(x\) is in radians. A candidate who has their calculator in degrees, or who differentiates \(\sin(x^{\circ})\) as if it were \(\sin x\), gets a gradient wrong by a factor of \(\tfrac{\pi}{180}\). The other reliable slip is the negative index: rewrite \(\dfrac{1}{x^2}\) as \(x^{-2}\) before differentiating, or the power rule has nothing to act on.
Differentiate \(y=3x^4-\dfrac{5}{x^2}+2\sqrt{x}+e^{2x}+\sin 3x+\ln x\).
Rewrite the awkward terms as powers first: \(-\dfrac{5}{x^2}=-5x^{-2}\) and \(2\sqrt{x}=2x^{1/2}\). Now go term by term. \(3x^4\to 12x^3\). \(-5x^{-2}\to -5(-2)x^{-3}=10x^{-3}=\dfrac{10}{x^3}\). \(2x^{1/2}\to 2\cdot\tfrac12 x^{-1/2}=x^{-1/2}=\dfrac{1}{\sqrt{x}}\). \(e^{2x}\to 2e^{2x}\). \(\sin 3x\to 3\cos 3x\). \(\ln x\to \dfrac{1}{x}\). Assembling:
\[\frac{dy}{dx}=12x^3+\frac{10}{x^3}+\frac{1}{\sqrt{x}}+2e^{2x}+3\cos 3x+\frac{1}{x}.\]
Every mark here is method: the examiner is checking the \(nx^{n-1}\) shift on the fractional and negative powers and the \(k\)-factor coming down from \(e^{2x}\) and \(\sin 3x\).
MechanismTurning points and optimisation — what the gradient being zero tells you
A stationary point is where \(f'(x)=0\): the tangent is horizontal. There are three kinds — local maximum, local minimum and stationary point of inflection — and you separate them with the second derivative. If \(f'(x_0)=0\) and \(f''(x_0)<0\), the gradient is falling through zero, so you have a maximum; if \(f''(x_0)>0\), the gradient is rising through zero, a minimum. If \(f''(x_0)=0\) the test is inconclusive and you fall back on checking the sign of \(f'\) just either side. Away from stationary points, \(f'(x)>0\) means the function is increasing and \(f'(x)<0\) means it is decreasing — the language every 'find the range of values of \(x\) for which \(f\) is increasing' question wants.
This is the engine of optimisation: the biggest volume, the least material, the maximum profit. The recipe never changes — write the quantity to be optimised as a function of one variable (using any constraint to eliminate the others), differentiate, set the derivative to zero, solve, then confirm max or min with the second derivative. The confirmation line is not optional politeness; it is a marked step.
A drinks can must hold \(330\) ml, i.e. \(330\) cm\(^3\). Find the radius that uses the least aluminium.
The can is a cylinder: volume \(V=\pi r^2 h=330\), surface area \(A=2\pi r^2+2\pi r h\) (two ends plus the curved side). Use the constraint to eliminate \(h\): \(h=\dfrac{330}{\pi r^2}\), so \(A=2\pi r^2+2\pi r\cdot\dfrac{330}{\pi r^2}=2\pi r^2+\dfrac{660}{r}\). Differentiate and set to zero: \(\dfrac{dA}{dr}=4\pi r-\dfrac{660}{r^2}=0\Rightarrow r^3=\dfrac{660}{4\pi}=52.5\Rightarrow r\approx 3.74\) cm. Confirm it is a minimum: \(\dfrac{d^2A}{dr^2}=4\pi+\dfrac{1320}{r^3}>0\) for all \(r>0\), so yes. The matching height is \(h=\dfrac{330}{\pi(3.74)^2}\approx 7.49\) cm — almost exactly \(2r\), so the material-minimising can is as tall as it is wide. Real cans are noticeably taller than this, which is a modelling point worth banking: the model ignores the thicker double-seamed ends, printing, and how a hand grips the can, so the true optimum is not the geometric one.
MechanismProduct, quotient and chain rules — differentiating combinations
The standard list handles single terms; three rules handle how terms are glued together. The chain rule differentiates a function of a function: if \(y=g(u)\) and \(u=h(x)\), then \(\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}\). So \((2x^2+1)^5\) differentiates to \(5(2x^2+1)^4\cdot 4x=20x(2x^2+1)^4\) — differentiate the outside, multiply by the derivative of the inside. The product rule: if \(y=uv\) then \(\dfrac{dy}{dx}=u'v+uv'\). The quotient rule: if \(y=\dfrac{u}{v}\) then \(\dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^2}\) — and the order of that numerator matters, because the minus sign makes it non-commutative.
The chain rule also unlocks two whole techniques. Implicit differentiation handles curves not written as \(y=\ldots\), such as \(x^2+y^2=25\): differentiate every term with respect to \(x\), remembering that any \(y\)-term carries a \(\dfrac{dy}{dx}\) by the chain rule, then rearrange. Parametric differentiation handles curves given as \(x=f(t),\ y=g(t)\): \(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\). And connected rates of change — the classic 'water pouring into a cone' problem — is just the chain rule chaining time to whatever else is moving.
A spherical balloon is inflated at a constant \(50\) cm\(^3\) per second. How fast is its radius increasing when \(r=5\) cm?
The unknown is \(\dfrac{dr}{dt}\); you are given \(\dfrac{dV}{dt}=50\). Link them through the volume \(V=\tfrac{4}{3}\pi r^3\), whose derivative is \(\dfrac{dV}{dr}=4\pi r^2\). The chain rule connects the three: \(\dfrac{dV}{dt}=\dfrac{dV}{dr}\cdot\dfrac{dr}{dt}\), so \(\dfrac{dr}{dt}=\dfrac{dV/dt}{dV/dr}=\dfrac{50}{4\pi r^2}\). At \(r=5\): \(\dfrac{dr}{dt}=\dfrac{50}{4\pi(25)}=\dfrac{50}{100\pi}=\dfrac{1}{2\pi}\approx 0.159\) cm/s. Notice the radius creeps ever more slowly as the balloon grows, because the \(r^2\) in the denominator keeps rising — the same \(50\) cm\(^3\) has to spread over a larger and larger surface.
CaseTangents and normals to parametric and implicit curves
Once you can find a gradient, a tangent and a normal are one substitution away. At a point with gradient \(m\), the tangent is the line through that point with gradient \(m\), written \(y-y_1=m(x-x_1)\). The normal is perpendicular to the curve there, so its gradient is \(-\dfrac{1}{m}\) — the negative reciprocal. That single relationship is where most marks are won or lost: students find \(m\) correctly and then use it for the normal too.
For a curve given implicitly, differentiate to get \(\dfrac{dy}{dx}\) in terms of both \(x\) and \(y\), then substitute the point's coordinates to get a number. For a curve given parametrically, use \(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\) and substitute the relevant value of \(t\). The only extra care is that a parametric point is defined by a \(t\)-value, so you evaluate the gradient at that \(t\), and you find the actual \((x,y)\) by putting the same \(t\) back into the parametric equations.
Find the tangent and normal to the curve \(x=t^2,\ y=2t\) at the point where \(t=2\).
Differentiate each parametric equation: \(\dfrac{dx}{dt}=2t\) and \(\dfrac{dy}{dt}=2\). So \(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{2}{2t}=\dfrac{1}{t}\). At \(t=2\) the gradient is \(m=\tfrac12\), and the point is \((x,y)=(2^2,\ 2\times 2)=(4,4)\). Tangent: \(y-4=\tfrac12(x-4)\), i.e. \(y=\tfrac12 x+2\). Normal: its gradient is \(-\dfrac{1}{1/2}=-2\), so \(y-4=-2(x-4)\), i.e. \(y=-2x+12\). As a check on the implicit route, the same curve satisfies \(x=\big(\tfrac{y}{2}\big)^2\), i.e. \(y^2=4x\); differentiating implicitly gives \(2y\dfrac{dy}{dx}=4\), so \(\dfrac{dy}{dx}=\dfrac{2}{y}=\dfrac{2}{4}=\tfrac12\) at \((4,4)\) — the two methods agree, as they must.
CaseBuilding a differential equation from a sentence
The final skill runs differentiation as a modelling tool: translating a description of how a rate behaves into an equation involving a derivative. The trigger phrase is almost always 'the rate of change of … is proportional to …'. 'Proportional to' becomes '\(=k\times\)', and a rate of change with time becomes \(\dfrac{d}{dt}\). So 'a population grows at a rate proportional to its current size \(P\)' becomes \(\dfrac{dP}{dt}=kP\), and 'a hot drink cools at a rate proportional to how much hotter it is than the \(20^{\circ}\)C room' becomes \(\dfrac{d\theta}{dt}=-k(\theta-20)\) — Newton's law of cooling, with the minus sign encoding that a hotter object loses heat.
The sign is the whole art. If the quantity is decaying (a radioactive sample, a draining tank, a cooling coffee), the rate is negative when the quantity is above its resting value, so a minus sign appears. If it is growing, the constant is positive. You are only constructing the equation here; actually solving \(\dfrac{dP}{dt}=kP\) to get \(P=P_0e^{kt}\) is the job of integration in Chapter 8 — but you cannot solve what you cannot set up, and setting up is where the applied-modelling marks live.
A tank drains so that the rate at which its water depth \(h\) falls is proportional to the square root of the current depth (Torricelli's law, roughly true for a small hole). Write the differential equation, and say what the constant's sign must be.
'The rate at which the depth falls' is \(-\dfrac{dh}{dt}\) if we take \(\dfrac{dh}{dt}\) as the (negative) rate of change of depth; 'proportional to \(\sqrt{h}\)' contributes \(k\sqrt{h}\). Since the depth is decreasing, \(\dfrac{dh}{dt}\) is negative, so the model is \(\dfrac{dh}{dt}=-k\sqrt{h}\) with \(k>0\). Read it back to check it makes sense: when \(h\) is large the tank drains fast, when \(h\) approaches zero the rate approaches zero — the last of the water dribbles out slowly, exactly as a real tap-off does. That sanity check, comparing the equation's behaviour to the physical story, is what turns a guessed equation into a defensible model.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Differentiation questions are marked method-first, so never jump to the answer. On a 'from first principles' question you must show the \(\frac{f(x+h)-f(x)}{h}\) quotient, the algebraic simplification, and the words 'as \(h\to 0\)' — a bald correct derivative scores almost nothing, because the whole point is the limit. When you differentiate, rewrite roots and reciprocals as powers (\(\sqrt{x}=x^{1/2}\), \(\frac{1}{x^2}=x^{-2}\)) before applying \(nx^{n-1}\), and keep your calculator in radians for anything trigonometric.
For tangents and normals, write the gradient function, substitute the point to get a number, and state clearly which gradient is the tangent's and which is the normal's negative reciprocal — label them, because muddling the two is the examiner's favourite trap. On stationary-point questions, solving \(f'(x)=0\) earns the first marks but the classification via \(f''\) is separately credited, so always add that line even when the shape is 'obvious'. In optimisation, use the constraint to reduce to one variable before differentiating, and finish with the second-derivative confirmation. For connected rates and parametric or implicit work, write the chain-rule link explicitly (\(\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}\)); the structural line is where the method marks sit, and it also stops you inverting a fraction by accident.