Calculus I › Differentiation rules › full formula sheet
The derivative of sec x
Secant’s derivative is the product sec x tan x — the quotient rule on 1/cos, and not to be confused with tangent’s sec².
Notation: sec x = 1/cos x. d/dx[sec(u)] = sec(u) tan(u)·u′. sec is undefined at π/2 + kπ — so is its derivative.
Before this lesson: Quotient rule · Derivative of cos x
Where it comes from
Secant is sec x = 1/cos x. Quotient rule with f = 1 (f′ = 0), g = cos x (g′ = −sin x):
Before reading on: sec x has its minimum value 1 at x = 0. What must its slope be there — and which candidate, sec x tan x or sec²x, gives exactly that?
The graph confirms it’s a product, not sec²: sec x has its minimum (value 1) at x = 0, so its slope there is 0. sec(0) tan(0) = 1·0 = 0 ✓, but the sec² guess gives sec²(0) = 1 ✗. Dead on arrival — and that test also separates it from tangent’s derivative.
Derivation
The derivation is the quotient computation above — stated as a clean proof:
Memory shape: d/dx [tan x] = sec²x (square, no tan factor) vs. d/dx [sec x] = sec x tan x (product, no square). The x = 0 test — slopes 1 vs. 0 — tells them apart forever.
How to use it
The procedure:
- Plain sec x → sec x tan x. A product — keep both factors.
- Something inside? Chain: d/dx [sec(u)] = sec(u) tan(u)·u′. Example: d/dx [sec(3x)] = 3 sec(3x) tan(3x).
- Products need the product rule: x² sec x differentiates as 2x sec x + x² sec x tan x.
- Check at 0: sec has a minimum at x = 0, so the derivative there is 0 — if your formula gives nonzero at 0, something’s off.
Judgment calls
sec²x means (sec x)² — chain: 2 sec x·sec x tan x = 2 sec²x tan x. Don’t square the product: sec x tan x is already the answer — squaring it is a different (wrong) function.
Worked examples
Four problems, easiest first. Keep both factors of the product.
Example 1 — d/dx [2 sec x]
- Pull out the 2: = 2·d/dx [sec x].
- Secant rule: = 2 sec x tan x = 2 sec x tan x.
Your turn: d/dx [−3 sec x]
Answer: −3 sec x tan x
Pull out the −3: −3·d/dx [sec x] = −3 sec x tan x. Secant’s answer is a product — keep both factors.
Example 2 — d/dx [sec(3x)] (chain)
- Layers: outer sec u, inner u = 3x.
- Outside: sec(3x) tan(3x). Inside’s derivative: 3.
- Multiply: = 3 sec(3x) tan(3x).
Your turn: d/dx [sec(4x)] (chain)
Answer: 4 sec(4x) tan(4x)
Layers: outer sec u, inner u = 4x. Outside: sec(4x) tan(4x). Inside’s derivative: 4. Multiply: 4 sec(4x) tan(4x).
Example 3 — d/dx [x² sec x] (product)
- Structure: multiplied — product rule: f = x², g = sec x.
- Derivatives: f′ = 2x, g′ = sec x tan x.
- Assemble: = 2x sec x + x² sec x tan x = x sec x (2 + x tan x).
Your turn: d/dx [x sec x] (product)
Answer: sec x (1 + x tan x)
Product rule: f = x, g = sec x. f′ = 1, g′ = sec x tan x. Assemble: sec x + x sec x tan x = sec x (1 + x tan x).
Example 4 — the minimum test: d/dx [sec x] at x = 0
- Formula: sec(0) tan(0) = 1 · 0 = 0.
- Geometry: sec x = 1/cos x bottoms out at (0, 1) — flat tangent, zero slope. Matches ✓
- The impostor: sec²(0) = 1 would claim slope 1 at a minimum — impossible. Test passed ✓
Your turn: The extremum test: d/dx [sec x] at x = π
Answer: 0
Formula: sec(π) tan(π) = (−1)·0 = 0. Geometry: sec x = 1/cos x peaks at (π, −1) — flat tangent, zero slope. The impostor sec²(π) = 1 would claim slope 1 at an extremum — impossible.
Memorization tips
- Product, not square: sec x tan x — two factors. Tangent’s sec² is the square. Different shapes, different rules.
- The 0-test: sec bottoms out at x = 0, so its slope is 0 there. sec(0) tan(0) = 0 ✓; sec²(0) = 1 ✗.
- Quotient in disguise: sec = 1/cos. Blank? Re-derive: [(0)(cos) − (−sin)]/cos² = sec tan in 30 seconds.
- Watch the double negative: −1·(−sin x) = +sin x — the plus is load-bearing. It’s why secant’s rule has no minus.
- Chains multiply: sec(3x) → 3 sec(3x) tan(3x). Keep the inside intact inside both factors.
- Domain travels: sec dies at π/2 + kπ, and so does sec tan. Never evaluate where sec is undefined.
Final challenge
Five mixed questions — product vs. square, 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 sec x?
d/dx[sec x] = sec x·tan x — a product, not a square. It’s the quotient rule on sec x = 1/cos x: [(0)(cos) − 1·(−sin)]/cos² = sin/cos² = sec x·tan x.
How do I avoid confusing it with tangent’s derivative?
d/dx[tan x] = sec²x (square, no tan factor) vs. d/dx[sec x] = sec x·tan x (product, no square). The x = 0 test: tan’s slope is 1, sec’s slope is 0 (it’s a minimum).
Why is there no minus sign?
The quotient’s numerator is (0)(cos x) − 1·(−sin x) = +sin x — the double negative makes it positive. Unlike the cofunction cot and csc, secant’s rule has no minus.
What is d/dx[sec(3x)]?
3sec(3x)·tan(3x). Chain rule: sec(3x)·tan(3x) for the outside, times the inside’s derivative 3.
What is d/dx[sec²x]?
2sec²x·tan x. That’s (sec x)² — chain rule: 2sec x times secant’s own derivative sec x·tan x.
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