A Level Maths · Topic guide

Pure: Differentiation

Differentiation is the process of finding the gradient function (derivative) of a curve, giving the rate of change or gradient at any point on it. A Level Maths covers differentiating polynomials, trig, exponential and log functions using the chain, product and quotient rules, then applying derivatives to find tangents, normals, stationary points and rates of change.

A LevelPureEdexcelAQAOCRWJEC

Before you start

Make sure you're comfortable with these topics first:

Method

  1. Differentiate each term of a polynomial using the rule ax^n -> anx^(n-1), rewriting roots and fractions as powers of x first if needed.
  2. For a composite function such as (expression)^n, sin(expression) or e^(expression), use the chain rule: differentiate the outer function, then multiply by the derivative of the inner expression.
  3. For a product of two functions u and v, use the product rule: dy/dx = u'v + uv'. For a quotient u/v, use the quotient rule: dy/dx = (u'v - uv')/v^2.
  4. Learn the standard derivatives: sin(x) -> cos(x), cos(x) -> -sin(x), e^x -> e^x, and ln(x) -> 1/x, applying the chain rule whenever the argument is not simply x.
  5. To find a tangent or normal at a point, substitute the x-value into the derivative to get the gradient, then use y - y1 = m(x - x1) with the tangent gradient m, or the negative reciprocal of m for the normal.
  6. To find and classify stationary points, set dy/dx = 0 and solve for x, then use the second derivative (positive means minimum, negative means maximum) or check the sign of the gradient either side.

Worked example

The curve C has equation y = (4x - 1)^3. Find dy/dx, and hence find the gradient of C at the point where x = 1.

  1. Use the chain rule: differentiate the outer power first, treating (4x-1) as a single block: dy/dx = 3(4x-1)^2 x (derivative of 4x-1).
  2. The derivative of 4x - 1 is 4.
  3. Combine: dy/dx = 3(4x-1)^2 x 4 = 12(4x-1)^2.
  4. Substitute x = 1: dy/dx = 12(4(1)-1)^2 = 12(3)^2 = 12 x 9.
  5. Final answer: gradient = 108.

Practice questions

Try each question, then tap to reveal the answer.

Q1Differentiate y = 5x^3 - 2x^2 + 7.Show answer

Answer: dy/dx = 15x^2 - 4x.

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Q2Find dy/dx for y = 3sqrt(x) - 4/x.Show answer

Answer: dy/dx = (3/2)x^(-1/2) + 4x^(-2), i.e. 3/(2sqrt(x)) + 4/x^2.

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Q3Find dy/dx for y = (5x + 2)^4 using the chain rule.Show answer

Answer: dy/dx = 20(5x + 2)^3.

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Q4Find dy/dx for y = x^2 * e^(3x) using the product rule.Show answer

Answer: dy/dx = e^(3x)(2x + 3x^2).

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Q5Find the coordinates of the stationary point of y = x^2 - 6x + 5, and state whether it is a maximum or minimum.Show answer

Answer: (3, -4), a minimum (since d^2y/dx^2 = 2 > 0).

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Q6Find dy/dx for y = ln(x)/x^2 using the quotient rule, as a single fraction.Show answer

Answer: dy/dx = (1 - 2ln(x))/x^3.

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Exam-style questions

Written in the style of a A Level Maths exam paper, with a full mark scheme.

Q1[3 marks]

Find dy/dx for y = 4x^3 - 3/x^2 + 2sqrt(x), for x > 0.

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Q2[5 marks]

The curve C has equation y = (2x - 3)^5. Find the equation of the tangent to C at the point where x = 2, giving your answer in the form y = mx + c.

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Q3[6 marks]

The curve C has equation y = x^3 - 6x^2 + 9x + 2. Find the coordinates of the stationary points of C, and use the second derivative to determine the nature of each one.

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