Math / Helper companion / Trig derivatives
Uncut · book companion · ~86–95
Spend the slope tool on waves — then chain
Derivatives of sin & cos
You already have the slope tool from the intro climb. Apply it to the height and sideways of a turning radius — sin and cos — then open compositions: outer slope times inner slope. This rung must use the derivative Tool; it is not a brand-new topic dump of trig formulas. (derivatan av sin · cos · kedjeregeln)
V1 — Slope of sin via the limit tool.
Builds on: f′ as limit of average rates / derivative function (derivative-intro).
Tool gained: (sin x)′ = cos x (x in radians for the clean form).
- Principle. The same zoom that turned Δy/Δx into f′ works on the height of a unit turn: the instantaneous rate of height-change is the sideways coordinate.
- Model. d/dx [sin x] = cos x. (Use radians so the limit identity matches; degrees need a scale factor — a later transfer.)
- Prediction / Answer. At x = π/2, sin has slope cos(π/2) = 0 — crest. At x = 0, slope = 1 — steepest climb through the origin.
V2 — Slope of cos from the same tool.
Builds on: (sin x)′ = cos x and the derivative limit (V1 / intro).
Tool gained: (cos x)′ = −sin x.
- Principle. Sideways change on the same unit tip is not a new cult — apply the slope tool again (or shift: cos x = sin(x+π/2) and use V1).
- Model. d/dx [cos x] = −sin x.
- Prediction / Answer. At x = 0, cos has slope 0 (flat top of sideways). At x = π/2, slope = −1 — falling through zero height of cos.
V3 — Chain start: outer × inner′.
Builds on: (sin)′/(cos)′ plus derivative of an inner function (intro power tool).
Tool gained: chain rule — (f(g(x)))′ = f′(g(x)) · g′(x); e.g. (sin(kx))′ = k cos(kx).
- Principle. A machine built from an outer map after an inner map: the output’s rate is outer’s slope at the inner value, times how fast the inner is changing.
- Model. (f ∘ g)′ = (f′ ∘ g) · g′. Special case: sin(kx) → k cos(kx); cos(g(x)) → −sin(g(x)) · g′(x).
- Prediction / Answer. y = sin(3x) → y′ = 3 cos(3x). Transfer: a wave compressed in time has steeper slopes by the same factor k.
1. Poor dump → rewrite
Book dump: a boxed list “(sin x)′ = cos x, (cos x)′ = −sin x, chain rule …” to chant. Our rewrite: each line spends the prior Tool — limit slope on sin, then cos, then outer×inner′ for compositions.
Weapons: (sin)′=cos → (cos)′=−sin → chain (f∘g)′ = f′(g)·g′. Product/quotient is the next climb, not a side dump here.
2. Hands-on — chain factor k
Compare slopes of sin(x) vs sin(kx) at x = 0 (both pass through 0 with slope k).
…
3. Fake → real
Fake
“Differentiate y = cos(5x).” No sensor, no reason.
Real
A tide gauge fits depth ≈ A cos(5t) (t in the model’s time unit). Vertical speed is −5A sin(5t). You need the chain factor 5 to size a pump that must match peak rate — same derivative, now a flow job.
Gate — four, all required
1. Reconstruct: (sin x)′ = cos x uses which prior Tool?
2. Why is chain after sin′/cos′ in the climb?
3. Predict: d/dx [cos(2x)] = ?
4. Transfer: y = sin(ωt), ω fixed. Peak |y′| equals…