Calculus (Extended Part Module 2): Limits, Derivatives and Integrals
Calculus (Extended Part Module 2)
The Calculus domain tests limits, differentiation, integration, and applications. HKDSE M2 questions ask you to choose the correct rule, justify a limit, and interpret a derivative or integral. Typical tasks include finding a tangent, locating a maximum, evaluating a definite integral, and using the result to find an area.
Core rules and concepts
- Limit: lim_{x→a} f(x) exists only if left- and right-hand limits agree; for rational functions, factor and cancel before substituting.
- Differentiation: product rule (uv)' = u'v + uv'; quotient rule (u/v)' = (u'v − uv')/v^2; chain rule dy/dx = (dy/du)(du/dx).
- Implicit differentiation: differentiate both sides with respect to x, treating y as a function of x; for example, d/dx(y^2) = 2y dy/dx.
- Applications: f'(x) = 0 gives stationary points; classify with f''(x) or a sign table. The tangent at x = a is y − f(a) = f'(a)(x − a).
- Integration: ∫x^n dx = x^(n+1)/(n+1) + C for n ≠ −1. For substitution, change dx as well as the integrand. A definite integral gives signed area; split intervals or take absolute values for total area.
Worked example
Find dy/dx when y = x^2 e^(3x).
Use the product rule with u = x^2 and v = e^(3x). Then u' = 2x and, by the chain rule, v' = 3e^(3x). Hence dy/dx = 2x e^(3x) + x^2(3e^(3x)) = x e^(3x)(2 + 3x). Reasoning: identify the product, differentiate each factor, then factor the result. A common error is differentiating e^(3x) as e^(3x) and forgetting the factor 3.
Traps that cost marks
- Cancelling x in a limit without stating x ≠ a.
- Reversing the quotient rule: the numerator is u'v − uv', not uv' − u'v.
- Omitting +C in an indefinite integral.
- In a definite integral with substitution, forgetting to change the limits or to substitute back.
- Treating f''(x) = 0 as sufficient for a point of inflection; a sign change is required.