§ 00 · Visual overview

See the slope.
Don't just memorize the rule.

Calculus is two questions asked of every curve: how fast is it changing, and how much has it piled up? Limits, derivatives, integrals, and the theorem that ties them together all live here as simulations you can drag, sweep, and replay. The formulas aren't calculus — they're the caption. The curve is calculus.

Tangent line → Slope machine → Riemann sums → Area = antiderivative →
a wandering curve its tangent line A tangent line surfs along the curve, forever reporting the local slope. That single picture is the whole idea of a derivative — it loops on its own.
§ 01 · Limits & the tangent line

A tangent is a secant that stopped waiting

You can't measure a slope "at an instant" directly — an instant has no length to divide by. So cheat: pick a second point h away, measure the ordinary two-point slope between them, then watch what happens as h shrinks toward zero. The number that slope approaches is the derivative.

slope = [f(a+h) − f(a)] / h  →  f′(a) as h → 0
f(x) = x³/3 − x secant through a and a+h tangent at a Drag h toward zero and watch the secant collapse onto the tangent.
Secant slope
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Tangent slope f′(a)
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Difference
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f(a)
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Notice the tangent slope f′(a) = a² − 1 never moves as you drag h — only the secant chases it, overshooting less and less. At a = 1 or a = −1 the tangent goes perfectly flat: those are exactly the hilltop and valley of the curve, the points calculus calls critical points.

§ 02 · Derivatives — the slope machine

Every curve hides a second curve

The derivative isn't one number — it's a whole new function, one slope value for every input x. A line's slope never changes; a parabola's slope grows steadily; sin(x)'s slope, remarkably, turns out to be cos(x). Pick a shape below and watch its "slope function" trace itself out live.

d/dx[xⁿ] = n·xn−1  ·  d/dx[sin x] = cos x  ·  d/dx[eˣ] = eˣ
f(x) f′(x) Drag the probe — the dot on f′ always reads the slope of the line touching f.
x
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f(x)
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f′(x) — slope
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Rule at play
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Try sin(x²): it isn't a basic shape, it's sin applied to another function, x². The chain rule says differentiate the outside, then multiply by the derivative of the inside — cos(x²) · 2x. Every "complicated" derivative is just several simple rules, chained.

§ 03 · Integrals — accumulating area

Area is just a limit of rectangles

Chop the region under a curve into thin vertical strips, pretend each one is a rectangle, and add up the areas. With a handful of strips it's a crude guess — but let the strips multiply and thin out, and the crude sum converges on one exact number: the definite integral.

∫ₐᵇ f(x) dx = limn→∞ Σ f(xᵢ*)·Δx
rectangles f(x) = 0.4x² − 2x + 5 Left / right / midpoint change which corner of the strip sets its height.
Riemann sum
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Exact value (FTC)
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Error
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n
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Watch the error column as n climbs: left/right endpoints shrink error roughly like 1/n, but midpoint and trapezoid — which average out the curve's bend — shrink it more like 1/n². Doubling n barely helps left-endpoint estimates; it helps midpoint enormously. That's why real numerical software almost never uses plain endpoints.

§ 04 · The Fundamental Theorem of Calculus

Area's derivative is the function itself

Define A(x) as the running area under f from 0 up to x. Nudge x forward a hair, and the area gains a thin sliver whose height is f(x) — so the rate the area grows, A′(x), is exactly f(x). Integration and differentiation, calculus's two halves, are inverse operations.

A(x) = ∫₀ˣ f(t) dt   ⇒   A′(x) = f(x)
f(t) = sin(t) + 1.5 shaded area = A(x) A(x) and its tangent Sweep x — the tangent slope on A always matches the height of f.
x
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f(x) = A′(x)
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A(x) — area so far
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Slope of A at x
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The last two readouts should always agree — f(x) and the slope of A(x) are the same number, measured two different ways. That agreement, for any continuous f, is the entire Fundamental Theorem. It's why you can compute an area (a sum of infinitely many rectangles) just by finding one antiderivative and subtracting two values.

§ 05 · Practice

Now you do the calculus

Each problem needs one idea from above. Hints reveal one step at a time — try before you peek, and check your work in the Evaluator.

1 · The power rule, plainly

Find the derivative of f(x) = 4x³ − 7x + 2, then evaluate it at x = 2.

Hint 1 — Differentiate term by term: d/dx[4x³] = 12x², d/dx[−7x] = −7, d/dx[2] = 0.
Hint 2 — So f′(x) = 12x² − 7. Plug in x = 2.
Answer — f′(2) = 12(4) − 7 = 41. Type 4x^3 - 7x + 2 into the Evaluator and probe x = 2 to confirm.

2 · Derivative from the definition

Using the limit definition, show that the derivative of x² is 2x — no shortcut rules allowed.

Hint 1 — Start from the definition: f′(x) = lim(h→0) [(x+h)² − x²] / h.
Hint 2 — Expand the numerator: (x+h)² − x² = 2xh + h². Divide by h: 2x + h.
Answer — As h → 0, 2x + h → 2x. Every power rule you've ever used is this same expand-and-cancel trick, done once and for all. See it live in § 01 with a different f.

3 · Rectangles vs. the shortcut

Estimate ∫₀⁴ 3x dx with 4 right-endpoint rectangles, then compute the exact value and compare.

Hint 1 — Each strip has width Δx = 1. Right endpoints are x = 1, 2, 3, 4, so heights are 3, 6, 9, 12.
Hint 2 — Sum: (3+6+9+12)×1 = 30. The exact area is a triangle: ½ × base × height = ½ × 4 × 12.
Answer — Exact = 24. Right-endpoints overshoot to 30 because 3x is increasing — every rectangle's right corner sits above the line. Midpoint would nail it exactly, since 3x is a straight line. Check the pattern in § 03.

4 · FTC without integrating

Let A(x) = ∫₀ˣ (t² + 1) dt. Find A′(3) without ever computing the integral.

Hint 1 — The Fundamental Theorem says A′(x) = f(x), the very function being integrated — no antiderivative needed.
Hint 2 — Here f(t) = t² + 1, so A′(x) = x² + 1.
Answer — A′(3) = 9 + 1 = 10. The theorem turns an infinite accumulation into a plug-in — that's the payoff explored in § 04.