amazon.co.ukFirst principles, the product, quotient and chain rules, tangents and normals, stationary points, optimisation, implicit differentiation and connected rates of change - A-Level pure maths calculus (Edexcel 9MA0 Topic 7, AQA 7357 Section G) revised with worked examples and mark-scheme logic.
Differentiation from first principles (Edexcel 9MA0 7.1; AQA 7357 G1; OCR A H640 11.01) defines the derivative as the limit of chord gradients: f'(x) = lim as h -> 0 of (f(x + h) - f(x))/h. Exam questions ask for specific functions - x^2, x^3, 1/x, x^2 + x - or a general polynomial, and the mark scheme awards marks for the correct expansion, the cancellation of h, and taking the limit correctly.
Line 1: (f(x + h) - f(x))/h = ((x + h)^2 - x^2)/h (M1: correct setup).
Line 2: = (x^2 + 2xh + h^2 - x^2)/h = (2xh + h^2)/h (M1: expand, cancel x^2).
Line 3: = 2x + h, valid for h is not 0 (A1: cancellation of h shown).
Line 4: As h -> 0, this tends to 2x, so d/dx(x^2) = 2x (A1: limit taken).
The power rule (Edexcel 9MA0 7.2; AQA 7357 G2) extends to all real n: d/dx(x^n) = n x^(n-1). It handles negative and fractional powers as readily as positive integers: sqrt(x) = x^(1/2) differentiates to (1/2)x^(-1/2) = 1/(2sqrt(x)), and 1/x = x^(-1) differentiates to -x^(-2) = -1/x^2. The A* habit: rewrite roots and reciprocals as powers before differentiating, and rewrite the answer in the form the question asks for. Sum and difference rules follow term by term, so polynomials differentiate mechanically - errors come from sign slips on negative powers, not from difficulty. Notation matters in proofs and applied questions alike: f'(x), dy/dx and d/dx of an expression all mean the same operation, but applied questions expect dy/dt style rates, and mixing notation mid-solution reads badly to examiners.
Three rules handle every composite or combined function (Edexcel 9MA0 7.5; AQA 7357 G2; OCR A H640 11.01/16.01). The product rule: d/dx(fg) = f'g + fg'. The quotient rule: d/dx(f/g) = (f'g - fg')/g^2 - note the order in the numerator, f' first. The chain rule: differentiate the outside function at the inside function, then multiply by the derivative of the inside: d/dx(f(g(x))) = f'(g(x)) x g'(x). With u = g(x), the chain rule reads dy/dx = dy/du x du/dx.
The standard derivatives to recall (Edexcel 9MA0 7.2): d/dx(e^x) = e^x; d/dx(ln x) = 1/x; d/dx(sin x) = cos x; d/dx(cos x) = -sin x; d/dx(tan x) = sec^2 x. With constants: d/dx(e^(3x)) = 3e^(3x), d/dx(sin(2x + 1)) = 2cos(2x + 1) - the chain rule supplies the internal factor.
Differentiate y = x^2 sqrt(2x + 1), simplifying your answer.
Line 1: u = x^2 and v = (2x + 1)^(1/2), so du/dx = 2x and dv/dx = (1/2)(2x + 1)^(-1/2) x 2 = (2x + 1)^(-1/2) (M1: product rule setup, chain on v).
Line 2: dy/dx = 2x(2x + 1)^(1/2) + x^2(2x + 1)^(-1/2) (M1: product rule applied).
Line 3: Common denominator: dy/dx = [2x(2x + 1) + x^2]/(2x + 1)^(1/2) = (5x^2 + 2x)/sqrt(2x + 1) (A1).
Line 4: Factorise: dy/dx = x(5x + 2)/sqrt(2x + 1) (A1: simplified form).
Rule choice is a decision, not a guess: two factors multiplied - product rule; a ratio - quotient rule; a function inside a function - chain rule; and combinations nest the rules inside each other. Writing u and v (or u and the inside function) explicitly earns the method mark even if the algebra later slips.
Differentiate y = sin(x)/x^2 and simplify your answer.
Line 1: u = sin(x), v = x^2; du/dx = cos(x), dv/dx = 2x (M1: setup).
Line 2: Quotient rule: dy/dx = (cos(x) x x^2 - sin(x) x 2x)/x^4 (M1: applied with f' first).
Line 3: Simplify: dy/dx = (x cos(x) - 2sin(x))/x^3 (A1: factor x cancelled).
The quotient rule is the product rule with a sign flipped and a denominator added - but mixing them up is common, so memorise the numerator order f'g - fg' with the derivative of the top function first.
The tangent to y = f(x) at (x1, y1) has gradient f'(x1) and equation y - y1 = f'(x1)(x - x1). The normal is perpendicular: gradient -1/f'(x1). The edge case worth knowing: when f'(x1) = 0, the tangent is horizontal and the normal is the vertical line x = x1 - no reciprocal exists, so write the vertical line directly.
Stationary points satisfy f'(x) = 0 (Edexcel 9MA0 7.4; AQA 7357 G3; OCR A H640 11.01). Classification uses the second derivative test: f'' < 0 at the point means a maximum (gradient decreasing through zero), f'' > 0 means a minimum. A point of inflection with f' = 0 needs f'' = 0 and a sign change in f' - the second derivative test is inconclusive there, and the sign of f' either side settles it.
Find the stationary points of y = x^3 - 3x^2 - 9x + 5 and determine their nature.
Line 1: dy/dx = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1) (M1: differentiate, factorise).
Line 2: dy/dx = 0 at x = 3 and x = -1 (A1: both stationary points found).
Line 3: d^2y/dx^2 = 6x - 6. At x = 3: 12 > 0, so a minimum (M1: second derivative test).
Line 4: At x = -1: -12 < 0, so a maximum (M1).
Line 5: y(3) = 27 - 27 - 27 + 5 = -22 and y(-1) = -1 - 3 + 9 + 5 = 10: minimum at (3, -22), maximum at (-1, 10) (A1: coordinates).
Always give the coordinates of stationary points, not just the x-values - mark schemes award the accuracy mark for the y-coordinate, and forgetting it is the most common drop in this question type.
Optimisation questions (Edexcel 9MA0 7.4; AQA 7357 G3; OCR A H640 16.01) ask for a maximum or minimum of a modelled quantity. The routine: (1) write the quantity as a function of one variable, using a constraint to eliminate the second variable; (2) differentiate and set the derivative to zero; (3) solve; (4) justify the nature with the second derivative or a sign check; (5) answer in context with units, checking the value lies in the sensible domain.
A closed rectangular box has a square base and volume 500 cm^3. Find its minimum surface area.
Line 1: Let the base side be x and the height h. Volume: x^2 h = 500, so h = 500/x^2 (M1: constraint used to eliminate h).
Line 2: Surface area: S = x^2 + 4xh = x^2 + 4x(500/x^2) = x^2 + 2000/x (M1: single-variable expression).
Line 3: dS/dx = 2x - 2000/x^2. Setting to zero: 2x^3 = 2000, so x = 10 (A1: critical value).
Line 4: d^2S/dx^2 = 2 + 4000/x^3, which is positive at x = 10, so a minimum (A1: justification).
Line 5: h = 500/100 = 5; S = 100 + 200 = 300 cm^2 (A1: answer in context).
Three discipline points: eliminate the extra variable before differentiating - differentiating a two-variable expression earns nothing; justify the nature of the stationary point explicitly (the second derivative line is a mark); and check the domain - a negative or zero base side is meaningless, and some optimisation problems have their maximum at an endpoint rather than a stationary point, so test the sensible endpoints when the model restricts the domain.
Implicit differentiation (Edexcel 9MA0 7.6; AQA 7357 G3; OCR A H640 16.01) handles curves not written as y = f(x). Differentiate every term with respect to x; for terms in y, the chain rule appends a dy/dx factor (d/dx of y^2 is 2y dy/dx; of xy is y + x dy/dx by the product rule). Then collect the dy/dx terms and solve.
A curve is defined by x^2 + 2xy + 3y^2 = 6. Find the gradient at the point (1, 1).
Line 1: 2x + (2y + 2x dy/dx) + 6y dy/dx = 0 (M1: implicit differentiation, product rule on xy).
Line 2: Collect: (2x + 6y) dy/dx = -(2x + 2y), so dy/dx = -(x + y)/(x + 3y) (A1: solved for dy/dx).
Line 3: At (1, 1): dy/dx = -(1 + 1)/(1 + 3) = -2/4 = -1/2 (A1: evaluated).
Connected rates of change apply the chain rule to rates: dy/dt = (dy/dx) x (dx/dt). For a sphere with radius growing at 0.5 cm/s: V = (4/3)(pi)r^3, so dV/dt = 4(pi)r^2 x dr/dt = 4(pi)(100)(0.5) = 200(pi) = 628 cm^3/s when r = 10 - differentiate first, then substitute, never the reverse.
Exam technique: (1) for from-first-principles questions, show the limit notation and the cancellation of h explicitly; (2) write u and v before applying the product or quotient rule; (3) give stationary point coordinates; (4) in optimisation, state the model and the constraint before differentiating, and justify the maximum or minimum; (5) in implicit work, the dy/dx factor on y-terms is the method mark. Also know the shapes: a positive cubic has an N-shape with max then min (or one stationary point of inflection when the roots of f' coincide), and a negative cubic reverses it - sketching before classifying catches sign errors. Board coverage: Edexcel 9MA0 Topic 7, AQA 7357 G1-G4, OCR A H640 11.01 and 16.01, WJEC/Eduqas AS/A2 Units, CCEA AS1/A21 - identical techniques, differing wording.