Calculus I › Differentiation rules › full formula sheet
The quotient rule
How to differentiate one function divided by another — with a numerator order you must get exactly right.
Notation: f is the top (numerator), g the bottom (denominator). Both differentiable, and g ≠ 0 wherever you use the rule.
Before this lesson: Product rule · Definition of the derivative
Where it comes from
The product rule handles multiplication; division needs its own rule. The tempting guess is (f/g)′ = f′/g′ — differentiate top and bottom separately. Kill it with f(x) = x², g(x) = x:
Before reading on: (x²/x)′ = 1 by simplifying first. What does the naive guess f′/g′ = 2x/1 predict at x = 5 — and can both be right?
The intuition: f/g = f · (1/g), so the quotient rule is secretly the product rule plus the chain rule on 1/g. Differentiating 1/g produces the minus sign and the g² — both features of the formula trace back to that one move, as the proof shows.
Derivation
Let F(x) = f(x)/g(x), with f, g differentiable and g(x) ≠ 0. Apply the definition and add zero in disguise — the same trick as the product rule’s proof:
Before reading on: f/g = f·(1/g), and d/dx [1/g] = −g′/g² by the chain rule. Where must that minus land in the quotient formula — and why can the order never be fg′ − f′g?
Where the minus comes from: d/dx [1/g] = −g′/g² (chain rule on g−1). The quotient rule’s minus sign and squared denominator are that computation, unpacked.
How to use it
The procedure, every time:
- Name top and bottom: f = numerator, g = denominator. (Order matters here — unlike the product rule, this rule is not symmetric.)
- Differentiate each separately: f′ and g′, fully.
- Chant it: “low d-high, minus high d-low, over low-low” — (f′g − fg′)/g². Say it as you write; the chant is the formula.
- Simplify the top. Factor where possible — most textbook answers collapse nicely.
Rewrite first, or rule first?
If the denominator is a lone power of x, rewrite as a product instead: 1/x² = x−2 differentiates in one line (−2x−3), while the quotient rule takes four. Reach for the quotient rule when top and bottom both genuinely depend on x — sin x/x, (x²+1)/(x²−1), ex/(x+2).
Worked examples
Four problems, easiest first. In each one, check the numerator order twice.
Example 1 — d/dx [x/(x+1)]
- Name: f = x (f′ = 1), g = x+1 (g′ = 1).
- Chant: low d-high minus high d-low, over low-low = [(1)(x+1) − x(1)] / (x+1)².
- Simplify top: (x+1 − x) = 1, so the answer is 1/(x+1)².
Your turn: d/dx [(x−1)/(x+2)]
Answer: 3/(x+2)²
f = x−1 (f′ = 1), g = x+2 (g′ = 1). Chant: [(1)(x+2) − (x−1)(1)]/(x+2)² = (x+2−x+1)/(x+2)² = 3/(x+2)².
Example 2 — d/dx [sin x / x]
- Name: f = sin x (f′ = cos x), g = x (g′ = 1).
- Chant: = [cos x · x − sin x · 1] / x².
- Tidy: (x cos x − sin x)/x². (Why minus? The sine’s growth fights the shrinking 1/x factor.)
Your turn: d/dx [cos x / x]
Answer: −(x sin x + cos x)/x²
f = cos x (f′ = −sin x), g = x (g′ = 1). Chant: [(−sin x)(x) − (cos x)(1)]/x² = −(x sin x + cos x)/x².
Example 3 — d/dx [(x²+1)/(x²−1)] (simplifies beautifully)
- Name: f = x²+1 (f′ = 2x), g = x²−1 (g′ = 2x).
- Chant: = [(2x)(x²−1) − (x²+1)(2x)] / (x²−1)².
- Expand top: (2x³ − 2x) − (2x³ + 2x) = −4x.
- Answer: −4x/(x²−1)². (Why so clean? The x³ terms annihilate — a sign you did it right.)
Your turn: d/dx [(x²−1)/(x²+1)]
Answer: 4x/(x²+1)²
f′ = 2x, g′ = 2x. Chant: [2x(x²+1) − (x²−1)2x]/(x²+1)² = (2x³+2x−2x³+2x)/(x²+1)² = 4x/(x²+1)². The x³ terms annihilate.
Example 4 — deriving tan′: d/dx [sin x/cos x]
- Name: f = sin x (f′ = cos x), g = cos x (g′ = −sin x).
- Chant: = [cos x·cos x − sin x·(−sin x)] / cos²x.
- Simplify: = (cos²x + sin²x)/cos²x = 1/cos²x = sec²x.
The lesson: the tan rule isn’t a new fact — it’s the quotient rule wearing a trig costume. Whenever you forget a trig derivative, re-derive it from sine and cosine.
Your turn: Deriving cot′: d/dx [cos x/sin x]
Answer: −csc²x
f = cos x (f′ = −sin x), g = sin x (g′ = cos x). Chant: [(−sin x)(sin x) − (cos x)(cos x)]/sin²x = −(sin²x+cos²x)/sin²x = −1/sin²x = −csc²x.
Memorization tips
- Chant it every time: “low d-high, minus high d-low, over low-low.” Students who chant don’t flip the numerator.
- f′g leads: the derivative of the top comes first. If your numerator starts with fg′, it’s backwards.
- Denominator always squared: g², never g. Forgetting the square is the second-most common slip.
- The x²/x test: (x²/x)′ = 1 kills f′/g′ (= 2x) in five seconds. Run it whenever you’re unsure.
- Rewrite lone powers: 5/x² = 5x−2 — one line, no quotient rule. Save the rule for two genuine functions.
- Expect cancellation: good quotient problems collapse (Example 3’s −4x). If your numerator stays ugly, recheck the algebra — don’t “simplify” by wishful thinking.
Final challenge
Five mixed questions — numerator order, simplification, and the tan connection. Score 5/5 and the rule is yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What is the quotient rule?
The quotient rule says (f/g)′ = (f′g − fg′)/g²: derivative-of-top times bottom, minus top times derivative-of-bottom, all over bottom squared. Chant: “low d-high minus high d-low, over low-low.”
Why isn’t (f/g)′ just f′/g′?
Test f(x) = x², g(x) = x: the quotient simplifies to x, whose derivative is 1 — but f′/g′ = 2x/1 = 2x. The limit proof forces the minus sign and the g², which the naive guess lacks.
Does the numerator order really matter?
Yes — (f/g)′ is not symmetric like the product rule. Writing (fg′−f′g)/g² flips the sign of the entire answer. f′g always leads.
When should I rewrite instead of using the quotient rule?
When the denominator is a lone power of x: 1/x³ = x^−3 differentiates in one line. Use the quotient rule when top and bottom both genuinely depend on x, like (x²+1)/(x²−1).
Where do the minus sign and g² come from?
From d/dx[1/g] = −g′/g² (chain rule on g^−1). Since f/g = f·(1/g), the quotient rule is the product rule plus that computation — the minus and the square are its fingerprints.
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