Differentiation and its applications
Differentiating powers, finding gradients, tangents and stationary points.
Review these first
Learning objectives
What you need to be able to do
Teacher-mapped phrasing — check against the official Cambridge syllabus for exact wording.
- 3.1.1Differentiate powers of x and use the chain, product and quotient rules.
- 3.1.2Find gradients, tangents, normals and stationary points, and determine their nature.
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Differentiation and its applications
Differentiation finds the gradient of a curve at a point — equivalently, the rate at which y changes with x.
The basic rule
If y = xⁿ then dy/dx = nxⁿ⁻¹: multiply by the power, then reduce the power by one.
y = x⁵→dy/dx = 5x⁴y = 3x²→dy/dx = 6xy = 7→dy/dx = 0(a constant has zero gradient)y = 4x→dy/dx = 4
The three rules
- Chain rule — for a function inside a function:
dy/dx = dy/du × du/dx. Fory = (3x + 1)⁵, differentiate the outside and multiply by the derivative of the inside:5(3x + 1)⁴ × 3 = 15(3x + 1)⁴. - Product rule — for
y = uv:dy/dx = u(dv/dx) + v(du/dx). - Quotient rule — for
y = u/v:dy/dx = (v(du/dx) − u(dv/dx)) / v². The order in the numerator matters, since subtraction is not commutative.
Tangents and normals
The gradient of the tangent at a point is the value of dy/dx there. The normal is perpendicular to the tangent, so its gradient is −1/(dy/dx). Method: differentiate, substitute the x-coordinate to get the gradient, then use y − y₁ = m(x − x₁).
Stationary points
At a stationary point the curve is momentarily flat, so dy/dx = 0. Solve that equation to find them. To determine the nature, use the second derivative d²y/dx²:
- Positive → minimum (curve bending upwards).
- Negative → maximum (curve bending downwards).
- Zero → inconclusive; examine the gradient either side.
Why it matters
Differentiation answers optimisation questions — the largest volume for a given amount of material, the minimum cost, the maximum height of a projectile — all of which are stationary point problems in disguise.
Think of it like this
The derivative is a speedometer for a curve: it tells you how fast y is changing at one instant, not over a stretch. A stationary point is the moment the speedometer reads zero — the top of a hill or the bottom of a valley.
Worked examples
Method, step by step
Find the coordinates and nature of the stationary point of y = x² − 6x + 5.
- 1Differentiate: dy/dx = 2x − 6.
- 2Set equal to zero: 2x − 6 = 0, so x = 3.
- 3Substitute into the original: y = 9 − 18 + 5 = −4, giving the point (3, −4).
- 4Second derivative: d²y/dx² = 2, which is positive, so it is a minimum.
Minimum at (3, −4)
Find the equation of the tangent to y = x³ at the point where x = 2.
- 1Differentiate: dy/dx = 3x².
- 2At x = 2, the gradient is 3(2²) = 12.
- 3The y-coordinate is y = 2³ = 8, so the point is (2, 8).
- 4Use y − y₁ = m(x − x₁): y − 8 = 12(x − 2), so y = 12x − 16.
y = 12x − 16
Common misconceptions
- Forgetting the chain rule's inner derivative. Differentiating `(3x + 1)⁵` as `5(3x + 1)⁴` misses the factor of 3.
- Using the tangent gradient for the normal. The normal gradient is the negative reciprocal.
- Assuming `d²y/dx² > 0` means a maximum. Positive means **minimum** — the curve is bending upwards like a valley.
In the exam
- Rewrite roots and fractions as powers before differentiating: `√x` becomes `x^(1/2)`, and `1/x²` becomes `x⁻²`.
- For any "find the maximum/minimum" question, set `dy/dx = 0`, solve, then justify the nature with the second derivative — the justification is usually a separate mark.