Calculus I › Differentiation rules › full formula sheet
The derivative of csc x
Cosecant’s derivative is negative csc x cot x — the quotient rule on 1/sin, with the cofunction minus intact.
Notation: csc x = 1/sin x. d/dx[csc(u)] = −csc(u) cot(u)·u′. csc is undefined at kπ — so is its derivative.
Before this lesson: Quotient rule · Derivative of sin x
Where it comes from
Cosecant is csc x = 1/sin x. Quotient rule with f = 1 (f′ = 0), g = sin x (g′ = cos x):
Before reading on: csc x = 1/sin x. The quotient rule will differentiate f = 1 on top — what survives in the numerator, and will the answer carry a minus?
The graph confirms both features. csc x has its minimum (value 1) at x = π/2, so the slope there is 0: −csc(π/2) cot(π/2) = −1·0 = 0 ✓. The no-minus guess −csc² would give −1 ✗ — dead on arrival. Product shape, minus sign: both verified.
Derivation
The derivation is the quotient computation above — stated as a clean proof:
Sec vs. csc: d/dx [sec x] = +sec x tan x but d/dx [csc x] = −csc x cot x. Same product shape; the “co-” in cosecant keeps its minus warning. Compare the proofs: secant’s −(−sin) cancels, cosecant’s −(cos) doesn’t.
How to use it
The procedure:
- Plain csc x → −csc x cot x. Minus first — it’s a cofunction.
- Something inside? Chain: d/dx [csc(u)] = −csc(u) cot(u)·u′. Example: d/dx [csc(2x)] = −2 csc(2x) cot(2x).
- Check at π/2: csc bottoms out there, so the derivative is 0 — if your formula gives nonzero at π/2, recheck.
- Don’t square it: the answer is the product csc cot, not csc² — squaring is cotangent’s territory.
Judgment calls
csc²x = (csc x)² chains: 2 csc x·(−csc x cot x) = −2 csc²x cot x. Sign discipline: every cofunction (cos, cot, csc) carries its minus — write it before the chain factor.
Worked examples
Four problems, easiest first. Minus first, both factors kept.
Example 1 — d/dx [4 csc x]
- Pull out the 4: = 4·d/dx [csc x].
- Cosecant rule: = 4·(−csc x cot x) = −4 csc x cot x.
Your turn: d/dx [−2 csc x]
Answer: 2 csc x cot x
Pull out the −2: −2·d/dx [csc x] = −2·(−csc x cot x) = 2 csc x cot x. Two minuses make a plus.
Example 2 — d/dx [csc(2x)] (chain)
- Layers: outer csc u, inner u = 2x.
- Outside: −csc(2x) cot(2x). Inside’s derivative: 2.
- Multiply: = −2 csc(2x) cot(2x).
Your turn: d/dx [csc(3x)] (chain)
Answer: −3 csc(3x) cot(3x)
Layers: outer csc u, inner u = 3x. Outside: −csc(3x) cot(3x). Inside’s derivative: 3. Multiply: −3 csc(3x) cot(3x).
Example 3 — d/dx [x csc x] (product)
- Structure: multiplied — product rule: f = x, g = csc x.
- Derivatives: f′ = 1, g′ = −csc x cot x.
- Assemble: = csc x + x·(−csc x cot x) = csc x (1 − x cot x).
Your turn: d/dx [x² csc x] (product)
Answer: x csc x (2 − x cot x)
Product rule: f = x², g = csc x. f′ = 2x, g′ = −csc x cot x. Assemble: 2x csc x − x² csc x cot x = x csc x (2 − x cot x).
Example 4 — the minimum test at x = π/2
- Formula: −csc(π/2) cot(π/2) = −1 · 0 = 0.
- Geometry: csc x = 1/sin x bottoms out at (π/2, 1) — flat tangent, zero slope. Matches ✓
- The impostor: −csc²(π/2) = −1 would claim slope −1 at a minimum — impossible. Product shape confirmed ✓
Your turn: The extremum test at x = 3π/2
Answer: 0
Formula: −csc(3π/2) cot(3π/2) = −(−1)·0 = 0. Geometry: csc x peaks at (3π/2, −1) — flat tangent. The impostor −csc²(3π/2) = −1 would claim slope −1 at an extremum — impossible.
Memorization tips
- Minus-first cofunction: “co-” means minus — cos, cot, csc. Write it before anything else.
- Product, not square: −csc x cot x has two factors. csc² is cotangent’s territory.
- The π/2 minimum test: csc bottoms out there — derivative 0. Kills the −csc² impostor (which gives −1).
- Sec vs. csc: +sec tan vs. −csc cot. Same shape; the “co-” keeps its minus. Compare the two proofs to see why.
- Quotient in disguise: csc = 1/sin. Blank? Re-derive: [(0)(sin) − cos]/sin² in 30 seconds.
- Chains multiply: csc(2x) → −2 csc(2x) cot(2x). Minus, both factors, chain factor — all three, every time.
Final challenge
Five mixed questions — the surviving minus, chains, and the minimum test. 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 csc x?
d/dx[csc x] = −csc x·cot x — a product with a minus. It’s the quotient rule on csc x = 1/sin x: [(0)(sin) − 1·cos]/sin² = −cos/sin² = −csc x·cot x.
Why does it have a minus but secant’s doesn’t?
Compare the quotient numerators: secant’s is (0)(cos) − 1·(−sin) = +sin (double negative cancels); cosecant’s is (0)(sin) − 1·(cos) = −cos (no rescue). The “co-” in cosecant is your warning.
How do I avoid squaring it into −csc²x?
The π/2 test: csc bottoms out at π/2 (slope 0); −csc(π/2)·cot(π/2) = 0 ✓ but −csc²(π/2) = −1 ✗. Product shape confirmed in five seconds.
What is d/dx[csc(2x)]?
−2csc(2x)·cot(2x). Chain rule: −csc(2x)·cot(2x) for the outside, times the inside’s derivative 2.
What is d/dx[csc²x]?
−2csc²x·cot x. That’s (csc x)² — chain: 2csc x times cosecant’s own derivative −csc x·cot x.
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