Sec 4 A-Math Differentiation notes

The derivative f′(x) is defined as the gradient of the tangent to y = f(x) at the point x: the limit of the chords (f(x + h) − f(x))/h as h shrinks to 0. The same number reads as a rate of change: if s(t) is a displacement, s′(t) is a velocity. The notations f′(x), dy/dx, and their second-derivative partners f″(x), d²y/dx² (the gradient of the gradient, or acceleration-type quantity) are all examinable and interchangeable.

Two immediate uses: the value of dy/dx at a point is the gradient of the tangent there; and the sign of dy/dx across an interval tells whether the curve climbs or falls. Everything else in the chapter is technique for computing dy/dx quickly and then exploiting it.

Constant multiples and sums differentiate term by term: d/dx(6x³ − 2x⁻² + 4x^½) = 18x² + 4x⁻³ + 2x^−½. The power rule hides roots and reciprocals in plain sight: 1/x³ = x⁻³ differentiates to −3x⁻⁴, and √x = x^½ to ½x^−½ (that is 1/(2√x)). Rewrite every term as xⁿ before differentiating — it prevents most sign slips.

The chain rule also writes as d/dx f(g(x)) = f′(g(x))·g′(x): differentiate the outside, leave the inside alone, then multiply by the derivative of the inside. It works identically with ln, sin and cos: d/dx ln(5x) = (1/(5x)) × 5 = 1/x, and d/dx cos 2x = −2 sin 2x. Keep the order straight in the product and quotient rules — v′u′ with the wrong owner is the classic error.

The recipe: (1) find the point by substituting into y = f(x); (2) evaluate f′ at that x for the tangent gradient; (3) the normal gradient is the negative reciprocal −1/f′(a); (4) write the line through the point with that gradient, tidying to the requested form. A tangent touches the curve; a normal crosses it at right angles.

A function is increasing where dy/dx > 0 and decreasing where dy/dx < 0. Boundaries between the regimes solve dy/dx = 0 — the stationary points (the tangent is horizontal there). Factorise dy/dx whenever possible: each factor gives a candidate x.

Classify each stationary point with the second derivative test: d²y/dx² > 0 (curve smiles) → minimum; d²y/dx² < 0 (curve frowns) → maximum. When d²y/dx² = 0 the test is silent — check the sign change of dy/dx through the point instead. If dy/dx does not change sign, the point is a stationary point of inflexion (as at x = 0 for y = x³).

For quantities that change with time, differentiate the connecting equation with respect to t, using the chain rule where needed: for a sphere V = (4/3)πr³, dV/dt = 4πr²·dr/dt. Given any two of the three rates and the current size, the third follows. Substitute the moment after differentiating — plugging in numbers first freezes the equation.

Max-min problems follow a fixed protocol: (1) write the quantity to optimise as a formula; (2) if it has two variables, eliminate one with the constraint; (3) differentiate and solve the resulting equation; (4) justify the nature with the second derivative (or a sign change); (5) answer the actual question asked, with units. Marks sit on the justification as much as on the number.