Differentiation and applications
Differentiation and applications tests the technique of differentiating standard, composite, product, quotient, implicit and parametric functions, and the use of derivatives to analyse curves: gradients, tangents and normals, stationary points, rates of change and optimisation.
Rules of differentiation
Standard results include d/dx(xⁿ) = nx^(n−1), d/dx(eˣ) = eˣ, d/dx(ln x) = 1/x, d/dx(sin x) = cos x and d/dx(cos x) = −sin x, with angles in radians. The chain rule states dy/dx = dy/du × du/dx. The product rule gives (uv)′ = u′v + uv′. The quotient rule gives (u/v)′ = (u′v − uv′)/v².
Implicit and parametric differentiation
When a relation is not given as y = f(x), differentiate every term with respect to x, remembering that d/dx(yⁿ) = n y^(n−1) dy/dx. For parametric curves x = f(t), y = g(t), the gradient is dy/dx = (dy/dt)/(dx/dt).
Stationary points and optimisation
Stationary points satisfy dy/dx = 0. The second derivative classifies them: d²y/dx² > 0 indicates a minimum, < 0 indicates a maximum, and = 0 is inconclusive, requiring a sign analysis of the gradient either side. For optimisation, form an expression in one variable, differentiate, solve, and confirm the nature of the stationary point.
Worked example
Find and classify the stationary points of y = x³ − 3x² + 4.
dy/dx = 3x² − 6x = 3x(x − 2), so stationary points occur at x = 0 and x = 2. Then d²y/dx² = 6x − 6, which is −6 at x = 0 (maximum, y = 4) and +6 at x = 2 (minimum, y = 0).
Traps that cost marks
- Forgetting the inner derivative in the chain rule, such as omitting the 3 in d/dx(e^(3x)).
- Reversing the order in the quotient rule's numerator.
- Dropping dy/dx terms in implicit differentiation.
- Differentiating trigonometric functions while the calculator is in degrees.
- Assuming d²y/dx² = 0 means a point of inflection without checking.