Calculus I › Differentiation rules › full formula sheet
The derivative of cot x
Cotangent’s derivative is negative cosecant squared — the quotient rule on cos/sin, with a minus the graph demands.
Notation: csc x = 1/sin x. d/dx[cot(u)] = −csc²(u)·u′. cot is undefined at kπ — so is its derivative.
Before this lesson: Quotient rule · Derivative of sin x · Derivative of cos x
Where it comes from
Cotangent is cot x = cos x/sin x — another quotient in disguise. Run the chant “low d-high minus high d-low, over low-low”:
Before reading on: at x = π/2, cot crosses zero heading down. Positive or negative slope — and does that match tan’s derivative or oppose it?
The graph confirms the minus: at x = π/2, cot crosses zero heading down — slope −1. −csc²(π/2) = −1 ✓, while the no-minus guess +csc² gives +1 ✗. Dead on arrival.
Derivation
The derivation is the quotient computation above — stated cleanly, plus the alternate form:
Mirror of tan: tan′ = +sec²x, cot′ = −csc²x. The cofunctions (cos, cot, csc) carry the minus — the “co-” is your sign warning.
How to use it
The procedure:
- Plain cot x → −csc²x. Minus first.
- Something inside? Chain: d/dx [cot(u)] = −csc²(u)·u′. Example: d/dx [cot(3x)] = −3 csc²(3x).
- Powers chain too: d/dx [cot²x] = 2 cot x·(−csc²x).
- Choose your form: −csc²x = −(1 + cot²x) — use the latter when the problem speaks cot.
Judgment calls
The “co-” rule: every cofunction derivative carries a minus — cos (−sin), cot (−csc²), csc (−csc cot). If it starts with “co”, expect a minus. Forgot it? Re-derive from cos/sin in 30 seconds.
Worked examples
Four problems, easiest first. Minus first, always.
Example 1 — d/dx [cot(3x)]
- Layers: outer cot u, inner u = 3x.
- Outside: −csc²(3x). Inside’s derivative: 3.
- Multiply: = −3 csc²(3x).
Your turn: d/dx [cot(5x)]
Answer: −5 csc²(5x)
Layers: outer cot u, inner u = 5x. Outside: −csc²(5x). Inside’s derivative: 5. Multiply: −5 csc²(5x). Minus first!
Example 2 — d/dx [x cot x] (product)
- Structure: multiplied — product rule: f = x, g = cot x.
- Derivatives: f′ = 1, g′ = −csc²x.
- Assemble: = cot x − x csc²x.
Your turn: d/dx [x² cot x] (product)
Answer: 2x cot x − x² csc²x
Product rule: f = x², g = cot x. f′ = 2x, g′ = −csc²x. Assemble: 2x cot x + x²(−csc²x) = x(2 cot x − x csc²x).
Example 3 — d/dx [cot²x]
- Rewrite: (cot x)² — outer u², inner cot x.
- Outside: 2 cot x. Inside’s derivative: −csc²x.
- Multiply: = −2 cot x csc²x.
Your turn: d/dx [cot³x]
Answer: −3 cot²x csc²x
Rewrite: (cot x)³ — outer u³, inner cot x. Outside: 3 cot²x. Inside’s derivative: −csc²x. Multiply: −3 cot²x csc²x.
Example 4 — sanity check at x = π/2
- Formula: −csc²(π/2) = −1/sin²(π/2) = −1/1 = −1.
- Geometry: near π/2, cot x ≈ −(x − π/2) — a line of slope −1. Matches ✓
- Alternate form: −(1 + cot²(π/2)) = −(1 + 0) = −1. Agrees ✓
Your turn: Sanity check at x = π/4
Answer: −2
Formula: −csc²(π/4) = −1/sin²(π/4) = −1/(1/2) = −2. Alternate form: −(1 + cot²(π/4)) = −(1 + 1) = −2. Agrees.
Memorization tips
- Cofunction → minus: cos, cot, csc all carry minuses. If it starts with “co”, write the minus first.
- Quotient in disguise: cot = cos/sin. Blank on the formula? Re-derive in 30 seconds — the chant plus Pythagoras.
- The π/2 test: cot falls through zero there (slope −1). −csc²(π/2) = −1 ✓; +csc² gives +1 ✗.
- Two forms: −csc²x = −(1 + cot²x). Use whichever the problem’s language suggests.
- Mirror of tan: tan′ = +sec², cot′ = −csc². Learn them as a pair — same shape, opposite signs.
- Domain travels: cot dies at kπ, and so does −csc². Never evaluate where cot is undefined.
Final challenge
Five mixed questions — the minus sign, the 1+cot² form, and chains. Score 5/5 and it’s 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 derivative of cot x?
d/dx[cot x] = −csc²x. It’s the quotient rule on cot x = cos x/sin x: [(−sin)(sin) − (cos)(cos)]/sin² = −1/sin² = −csc²x.
Why the minus sign?
The graph demands it: cot falls through zero at x = π/2 (slope −1), and −csc²(π/2) = −1 fits. In the proof, the minus comes from cosine’s derivative (−sin) leading the quotient’s numerator.
Is there another form?
Yes: −csc²x = −(1 + cot²x), from dividing sin²+cos² = 1 by sin². Handy when the answer should stay in cot-language.
What is d/dx[cot(3x)]?
−3csc²(3x). Chain rule: −csc²(3x) for the outside, times the inside’s derivative 3.
How do I remember the cofunction signs?
Every “co-” derivative carries a minus: cos → −sin, cot → −csc², csc → −csc·cot. The “co” is your minus warning.
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