Calculus
What Calculus tests
Calculus tests change and accumulation: limits, derivatives, tangents and normals, stationary points, optimisation, and definite integrals as area. The practice expects correct notation and interpretation of signs.
Core rules
- Derivative: f'(x) = lim as h→0 of (f(x+h) − f(x))/h; power rule d/dx x^n = nx^(n−1); d/dx sin x = cos x, d/dx cos x = −sin x, d/dx e^x = e^x, d/dx ln x = 1/x.
- Chain rule: d/dx f(g(x)) = f'(g(x))g'(x).
- Tangent at x = a: slope f'(a); normal slope −1/f'(a) when f'(a) ≠ 0.
- Stationary points: solve f'(x) = 0; classify with sign change or f''.
- Integration: ∫x^n dx = x^(n+1)/(n+1) + C for n ≠ −1; definite integral ∫_a^b f(x)dx = F(b) − F(a); area below the axis gives a negative integral.
Worked example
For f(x) = x^3 − 3x^2 + 2, find and classify the stationary points. f'(x) = 3x^2 − 6x = 3x(x − 2), so x = 0 or x = 2. f''(x) = 6x − 6. At x = 0, f''(0) = −6 < 0, so local maximum; at x = 2, f''(2) = 6 > 0, so local minimum. Values: f(0) = 2, f(2) = −2.
Traps
Forgetting +C for indefinite integrals; evaluating F(a) − F(b) instead of F(b) − F(a); solving f(x) = 0 when asked for stationary points; not checking endpoints in optimisation; treating area below the axis as positive without splitting the interval.
Further analytical methods
Integration by parts follows from differentiating a product: ∫u dv=uv-∫v du. Choose u so that its derivative simplifies the remaining integral. In substitution, transform the differential and either change definite bounds to the new variable or convert back before using the original bounds. A Maclaurin approximation uses derivatives at zero and needs a justified range or error estimate; a polynomial approximation is not an identity everywhere. For a separable differential equation, check constant solutions that dividing by a dependent-variable factor could otherwise lose.