Differentiation
Differentiation finds the gradient of a curve at any point. With one rule for powers of x you can find tangents, normals and turning points, which is the start of almost every calculus question in the course.
Part 1 of 3: Learn it
In short
- If y = axⁿ, then dy/dx = naxⁿ⁻¹.
- Write roots and fractions as powers before differentiating: √x = x^(1/2), 1/x² = x⁻².
- At a stationary point dy/dx = 0; the sign of d²y/dx² tells you if it is a maximum or minimum.
Where this is in your specification
Spec points: DfE G1 to G3 (AQA 7357 G1 to G3, Edexcel 9MA0 Pure topic 7, OCR A H240 1.07)
| Board | Topic: Differentiation |
|---|---|
| DfE content | G1-G4 |
| AQA 7357 | G1-G4 |
| Edexcel 9MA0 | Pure topic 7 |
| OCR H240 | 1.07 |
| Higher C847 76 | Calculus skills: differentiation, stationary points, optimisation |
Gradient and first principles
At any point, a curve has the same gradient as the tangent touching it at that point. Take a second point a distance h along; the gradient of the chord joining them gets closer to the tangent's gradient as h shrinks to 0.
For f(x) = x²: (x + h)² − x² = 2xh + h², divide by h to get 2x + h, and as h → 0 this tends to 2x.
Differentiating powers
- Multiply by the power, then reduce the power by one: 5x⁴ becomes 20x³.
- A constant on its own differentiates to 0; a term like 7x becomes 7.
- Differentiate a sum term by term. Expand brackets and split fractions first.
Tangents and normals
Find the y value and the gradient m at the point. The tangent is y − y₁ = m(x − x₁). The normal is perpendicular to the tangent, so its gradient is −1/m.
Stationary points
Solve dy/dx = 0 for x, then find y. Differentiate again: if d²y/dx² > 0 the point is a minimum, if d²y/dx² < 0 a maximum. If it is 0, test the gradient either side. A function is increasing where dy/dx > 0 and decreasing where dy/dx < 0.
Differentiate y = 4x³ − 3x² + 7x − 5.
Show the answer
dy/dx = 12x² − 6x + 7.
Part 2 of 3: See it worked
Worked examples
Example 1
Find the equation of the tangent to y = x² − 3x + 1 at the point where x = 4.
- y = 16 − 12 + 1 = 5, so the point is (4, 5)
- dy/dx = 2x − 3, which is 5 at x = 4
- y − 5 = 5(x − 4)
Answer: y = 5x − 15.
Example 2
Find and classify the stationary points of y = x³ − 12x + 3.
- dy/dx = 3x² − 12 = 0, so x² = 4 and x = 2 or x = −2
- x = 2: y = 8 − 24 + 3 = −13; x = −2: y = −8 + 24 + 3 = 19
- d²y/dx² = 6x: at x = 2 it is 12 (positive), at x = −2 it is −12 (negative)
Answer: Minimum at (2, −13) and maximum at (−2, 19).
Common mistakes
- Differentiating a term like 3/x² without rewriting it as 3x⁻² first.
- Using the gradient of the tangent for the normal instead of −1/m.
- Forgetting to find the y coordinate of a stationary point.
- Lowering a negative power the wrong way: x⁻² differentiates to −2x⁻³, not −2x⁻¹.
Differentiate y = √x.
Show the answer
y = x^(1/2), so dy/dx = ½x^(−1/2), which is 1/(2√x).
Part 3 of 3: Test yourself
Check yourself
Answer each one in your head or on paper first, then open it to check.
Differentiate y = 4x³ − 3x² + 7x − 5.
dy/dx = 12x² − 6x + 7.
Differentiate y = √x.
y = x^(1/2), so dy/dx = ½x^(−1/2), which is 1/(2√x).
The gradient of a tangent is 4. What is the gradient of the normal?
−1/4.
Jobs that use this
- Aerospace engineer (se abre en otra pestaña)
- Economist (se abre en otra pestaña)
- Data scientist (se abre en otra pestaña)
Each link opens the job profile on the National Careers Service (England). In the rest of the UK: My World of Work (Scotland), Careers Wales, nidirect careers (Northern Ireland).
Estos apuntes están en inglés porque siguen los programas de examen del Reino Unido.
Full lessons and marked practice for this course are coming soon to Brainlag Learn. See courses