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
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