A-Level Maths differentiation: rules, stationary points and tangents

Where differentiation sits in each specification

Cambridge students take Paper 2 (AS only) or Paper 3 (full A Level), not both. Check which route your school enters.

The key ideas

The derivative is the gradient of the tangent at a point. For y = x^n the derivative is nx^(n-1), and this works for any rational n, so first rewrite roots and fractions as powers: sqrt(x) = x^(1/2) and 4/x^2 = 4x^(-2). Constants multiply through and sums are differentiated term by term.

The chain rule handles a function inside a function: if y = f(g(x)) then dy/dx = f'(g(x)) × g'(x). In words, differentiate the outside, keep the inside, then multiply by the derivative of the inside. The product rule is d(uv)/dx = u dv/dx + v du/dx, and the quotient rule is d(u/v)/dx = (v du/dx - u dv/dx)/v^2. The standard results to know are d(e^x)/dx = e^x, d(ln x)/dx = 1/x, d(sin x)/dx = cos x, d(cos x)/dx = -sin x and d(tan x)/dx = sec^2 x, with x always in radians.

Applications follow from the meaning. At a stationary point dy/dx = 0; if d2y/dx2 is negative it is a maximum and if positive a minimum. The tangent at x = a has gradient f'(a); the normal is perpendicular, so its gradient is -1/f'(a). Connected rates of change use the chain rule with time: dA/dt = dA/dr × dr/dt. For implicit curves you differentiate both sides with respect to x, remembering that d(y^2)/dx = 2y dy/dx.

Worked example 1: find and classify the stationary points of y = x^3 - 6x^2 + 9x + 2

1. Differentiate: dy/dx = 3x^2 - 12x + 9.
2. Set dy/dx = 0 and factorise: 3(x^2 - 4x + 3) = 3(x - 1)(x - 3) = 0, so x = 1 or x = 3.
3. Find the y values: at x = 1, y = 1 - 6 + 9 + 2 = 6; at x = 3, y = 27 - 54 + 27 + 2 = 2.
4. Second derivative: d2y/dx2 = 6x - 12. At x = 1 it is -6, which is negative, so (1, 6) is a maximum. At x = 3 it is 6, which is positive, so (3, 2) is a minimum.
5. Answer: maximum at (1, 6) and minimum at (3, 2). A quick sketch of a positive cubic (up, down, up) confirms the maximum comes first.

Worked example 2: the tangent and normal to y = sqrt(2x + 5) at x = 2

1. Write as a power: y = (2x + 5)^(1/2). At x = 2, y = sqrt(9) = 3, so the point is (2, 3).
2. Chain rule: dy/dx = (1/2)(2x + 5)^(-1/2) × 2 = 1/sqrt(2x + 5).
3. At x = 2 the gradient is 1/sqrt(9) = 1/3.
4. Tangent: y - 3 = (1/3)(x - 2), which rearranges to x - 3y + 7 = 0. Check: (2, 3) gives 2 - 9 + 7 = 0.
5. Normal gradient is -1 ÷ (1/3) = -3, so the normal is y - 3 = -3(x - 2), that is y = -3x + 9.

Worked example 3 (Cambridge P2/P3, Edexcel P4): the gradient of x^2 + y^2 = xy + 7 at (3, 1)

1. Check the point lies on the curve: 9 + 1 = 10 and 3 × 1 + 7 = 10.
2. Differentiate every term with respect to x. Use the product rule on xy: 2x + 2y dy/dx = y + x dy/dx.
3. Collect the dy/dx terms on one side: 2y dy/dx - x dy/dx = y - 2x, so dy/dx (2y - x) = y - 2x.
4. So dy/dx = (y - 2x)/(2y - x).
5. At (3, 1): dy/dx = (1 - 6)/(2 - 3) = (-5)/(-1) = 5.

Common mistakes that cost marks

• Forgetting the chain rule factor: the derivative of (3x - 1)^5 is 15(3x - 1)^4, not 5(3x - 1)^4.
• Differentiating sqrt(x) or 1/x without first rewriting as x^(1/2) or x^(-1).
• Swapping the order in the quotient rule numerator, which flips the sign of the whole answer.
• Using degrees: d(sin x)/dx = cos x only when x is in radians.
• Stating 'maximum' or 'minimum' with no reason. The mark is for the second derivative value or a sign check either side.
• In implicit differentiation, writing d(y^2)/dx = 2y and forgetting the dy/dx, or not using the product rule on xy.
• Giving the normal the same gradient as the tangent.

Exam technique and how a tutor helps

Differentiation appears on almost every pure paper, often as the first part of a longer question that then asks for a tangent, a stationary point or an area. Show the derivative on its own line before substituting, because the derivative earns a method mark even if the arithmetic afterwards slips. When a question says 'show that', every step must be visible. When it says 'hence', use the result you just found rather than starting again.

Cambridge Paper 1 and Edexcel IAL papers allow a calculator, but the algebra of factorising dy/dx = 0 still has to be done by hand. Always state the coordinates of a stationary point in full, and when the question asks for the nature, write the second derivative value and the conclusion together.

In one-to-one lessons a tutor works through questions with the student on the shared whiteboard and identifies which rule breaks under pressure, usually the chain rule hidden inside a product or quotient. The fix is short drills on that rule, then past-paper questions where differentiation is one step inside a bigger problem, marked against the official mark scheme so the student sees exactly where method marks are awarded.

Self-check: can you do these?

• Differentiate y = 4/x^2 - 3sqrt(x). (Answer: -8x^(-3) - (3/2)x^(-1/2))
• Differentiate y = x^2 ln x. (Answer: 2x ln x + x)
• Differentiate y = sin x / x. (Answer: (x cos x - sin x)/x^2)
• The radius of a circle grows at 0.2 cm per second. How fast is the area growing when r = 5 cm? (Answer: 2π, about 6.28 cm^2 per second)
• Find the stationary point of y = x^2 - 8x + 3 and say whether it is a maximum or a minimum. (Answer: (4, -13), minimum)