Stop taking limits one point at a time — look up the basics, then combine them with sum, product, quotient, and the chain.
For three lessons you computed every slope the hard way — set up the difference quotient, let Δx shrink, take the limit, one point at a time. That was the right way to build the idea, but no one differentiates like that forever. Calculus has a toolbox: a short table of derivatives for the basic functions, and a handful of rules for snapping them together. Learn the table — (xⁿ)′ = n xⁿ⁻¹, (eˣ)′ = eˣ, (ln x)′ = 1/x, (sin x)′ = cos x — and learn how a sum, a product, a quotient, and a composite each differentiate. After that, finding f′ is bookkeeping, not a limit. The one factor everybody forgets is the inner derivative of the chain rule — so we will dwell on it.
Every slope you found from the definition was really one entry in a small lookup table. Here is the table — the seven derivatives that the rest of calculus is built on. Read each row as "the derivative of the left thing is the right thing."
| function f | derivative f′ | in words |
|---|---|---|
| c (a constant) | 0 | a flat line has slope 0 |
| xn | n xn−1 | drop the power, subtract one |
| ex | ex | its own slope everywhere |
| ax | ax ln a | ex is the case a = e (ln e = 1) |
| ln x | 1/x | steep near 0, flat far out |
| sin x | cos x | slope of sine is cosine |
| cos x | −sin x | note the minus sign |
You don't have to take the table on faith. Pick a basic function below and we overlay f and its f′. Watch how the green curve reports the slope of the blue one: where f flattens out, f′ crosses zero; where f rises steeply, f′ is high.
These are facts to know, the way you know your times tables. The rules in the next four sections never produce a brand-new basic derivative — they only tell you how to combine the ones in this table.
The easiest rule comes free from the way limits add. The derivative of a sum is the sum of the derivatives, and a constant multiplier just rides along:
(u ± v)′ = u′ ± v′ and (c·u)′ = c·u′
So you differentiate a polynomial term by term. Each xⁿ uses the power rule; each coefficient stays put. There is no special "polynomial rule" — it is just the table plus this one.
Differentiate f(x) = 3x⁴ − 5x² + 7x − 2.
Term by term: (3x⁴)′ = 12x³, (−5x²)′ = −10x, (7x)′ = 7, (−2)′ = 0. Add them:
f′(x) = 12x³ − 10x + 7.
At x = 1 that is f′(1) = 12 − 10 + 7 = 9 — the slope of the curve there.
The power rule is the workhorse, so let's dial the exponent and watch f and f′ move together. The widget builds f = xⁿ from a coefficient list and takes f′ exactly with the power rule — no numbers are read off the picture.
Here is where the table stops being enough. The derivative of a product is not the product of the derivatives — the tempting u′v′ is wrong. Think of a rectangle of width u and height v: when both sides grow a little, the new area comes from the height growing (u·v′) and from the width growing (u′·v). Add both contributions:
(u·v)′ = u′·v + u·v′
Let each factor change in turn, holding the other still, then sum. "Derivative of the first times the second, plus the first times the derivative of the second."
Differentiate f(x) = x·eˣ. Take u = x and v = eˣ, so u′ = 1 and v′ = eˣ.
(x·eˣ)′ = (1)(eˣ) + (x)(eˣ) = (1 + x)eˣ.
At x = 1: f′(1) = (1 + 1)·e = 2e ≈ 5.437. (A numerical check with the definition gives the same 5.437 — the rule is honest.)
Multiplying the derivatives is the classic slip: (x·eˣ)′ is not (1)(eˣ) = eˣ. If that were true, the slope at x = 1 would be e ≈ 2.718 — but the true slope is 2e ≈ 5.437. The cross-terms matter.
A quotient u/v differentiates with a partner rule. The denominator gets squared, and the numerator subtracts in a fixed order — "low d-high minus high d-low, over low squared":
(u/v)′ = u′·v − u·v′v²
The order in the numerator is not symmetric — swap the two products and you flip the sign of the whole derivative, so keep u′v first.
Differentiate f(x) = x ⁄ (x + 1). Here u = x (u′ = 1) and v = x + 1 (v′ = 1).
f′(x) = (1)(x+1) − (x)(1)(x+1)² = 1(x+1)² = 1 ⁄ (x + 1)².
At x = 2: f′(2) = 1 ⁄ 3² = 1/9 ≈ 0.111. Always positive — the curve never stops rising, which the formula tells you at a glance.
Most functions you meet are not basic at all — they are composites, one function fed into another: y = sin(x²), y = (2x + 1)⁵, y = e3x. There is an outer function and an inner function tucked inside it, like gears in a chain. The chain rule says: differentiate the outer (leaving the inner alone), then multiply by the derivative of the inner.
If y = f(g(x)), then y′ = f′(g(x)) · g′(x) — outer′ at the inner, times inner′.
Outer is sin( ), inner is g = x². So outer′ = cos( ) and inner′ = 2x.
y′ = cos(x²) · 2x = 2x·cos(x²). The ·2x is the inner derivative — drop it and you are wrong.
The number-one chain-rule error is writing sin(x²)′ = cos(x²) and stopping. That misses the ·2x. The inner derivative is the whole point of the chain rule: cos(x²) is not the slope; cos(x²)·2x is.
Build a composite below and the widget assembles f′ in two pieces — outer′ at the inner, then ·inner′ — and confirms the result against a direct numerical derivative at one x. When the rule and the machine agree, you've nailed it.
The hardest part of differentiating is not the rules — it is reading the function correctly. Before you write a single derivative, name the outermost structure: is this a sum, a product, a quotient, or a composite? That decides which rule fires first.
| looks like | structure | rule to use first |
|---|---|---|
| x³ − 2x + sin x | sum | term by term |
| x² · sin x | product | u′v + uv′ |
| (sin x) / x | quotient | (u′v − uv′)/v² |
| sin(x²), e3x, (2x+1)⁵ | composite | chain: outer′·inner′ |
Differentiate f(x) = x²·sin x. The outermost structure is a product (u = x², v = sin x), so the product rule fires first:
f′(x) = (x²)′·sin x + x²·(sin x)′ = 2x·sin x + x²·cos x.
Neither piece needed the chain rule here — but if the inner had been sin(x²), you'd have reached for the chain inside the product. Name the structure, then descend.
| tool | rule | remember |
|---|---|---|
| power | (xⁿ)′ = n xⁿ⁻¹ | drop the power, subtract one |
| exp / log | (eˣ)′ = eˣ, (ln x)′ = 1/x | eˣ is its own derivative |
| trig | (sin x)′ = cos x, (cos x)′ = −sin x | cosine carries the minus |
| sum | (u ± v)′ = u′ ± v′ | differentiate term by term |
| product | (uv)′ = u′v + uv′ | not u′v′ |
| quotient | (u/v)′ = (u′v − uv′)/v² | low d-high − high d-low, keep order |
| chain | [f(g(x))]′ = f′(g(x))·g′(x) | never forget the inner ·g′(x) |
With these seven rules you can differentiate almost anything you'll meet. Next, in 36.5, the sign of f′ becomes a spotlight on the shape of a curve — where it rises, where it falls, and where its peaks and valleys hide. (If you want to revisit how a derivative was born from a limit, look back at 36.3.)
Six questions to lock it in. Tap the answer you think is right.
This lesson assembles the standard differentiation rules — power, exponential, logarithmic, and trigonometric derivatives, plus the sum, product, quotient, and chain rules — the core computational toolkit of AP Calculus AB. It builds on the formal definition (HSF-IF.B.6, rate of change) by replacing limit computations with reusable formulas, and it foregrounds the two misconceptions that cost the most points: that (uv)′ equals u′v′, and that the chain rule's inner derivative can be skipped. Every derivative quoted in a figure or readout is computed by an exact polynomial-derivative routine or checked against a numerical derivative, so the pictures never assert a slope they cannot back up.