sec, cosec, and cot are just the reciprocals of cos, sin, and tan — so their derivatives drop out of the quotient rule applied to 1/(cosx), 1/(sinx), and cosx/sinx. All three results are in the formula booklet, but the patterns (especially the negative signs) are easy to muddle up under exam pressure. Always cross-check with the booklet before submitting.
📘 What you need to know
Definitions: sec x = 1/cos x; cosec x = 1/sin x; cot x = 1/tan x = cos x/sin x. Angles in radians for all calculus work.
The three formula-booklet derivatives: d/dx[sec x] = sec x tan x; d/dx[cosec x] = −cosec x cot x; d/dx[cot x] = −cosec² x.
Sign pattern: sec → positive; cosec → NEGATIVE; cot → NEGATIVE. The “co-” functions all pick up a minus sign (like cos and cot themselves).
Each derivative is a PRODUCT of two reciprocal trig functions (except cot, whose derivative is just −cosec²).
Chain rule for linear arguments: d/dx[sec(ax+b)] = a sec(ax+b) tan(ax+b) — the inner derivative pulls out as a factor.
Product and quotient rules still apply normally — derivatives of, say, x sec x use product rule with the formula-booklet result for d/dx[sec x].
Always double-check from the booklet before writing the final answer — examiners deduct marks if you confuse cosec with sec, or drop a negative sign.
Be careful with the arguments inside two trig functions: d/dx[cosec 3x] = −3 cosec 3x cot 3x, NOT −3 cosec x cot 3x. Both factors carry the same inner expression.
The three derivatives — and the pattern
Formula booklet — reciprocal trig derivativesddx[sec x] = sec x tan x | ddx[cosec x] = − cosec x cot x | ddx[cot x] = − cosec² x
The “co-” functions (cosec and cot) both pick up a minus sign — mirroring how cos and cot have negative derivatives in their families. Sec is the only positive one.
Why the proofs work: each derivative drops out of the quotient rule on the definition. For sec x = 1/cos x, the quotient rule gives sin x/cos² x, which splits as (1/cos x)(sin x/cos x) = sec x tan x. Same idea for cosec and cot.
🧭 Recipe — differentiate any reciprocal trig expression
Identify the base reciprocal trig function (sec, cosec, or cot) and the argument (just x, a linear ax+b, or something nonlinear like x²).
Quote the formula-booklet result for the base function: sec → sec·tan; cosec → −cosec·cot; cot → −cosec². Check the negative signs.
Apply chain rule if the argument is anything other than x: multiply by the derivative of the inner expression. For linear ax+b, the inner derivative is just a.
Use product or quotient rule on top of that if the expression involves x·sec x, cotx/x, secx·tanx, etc. — the formula-booklet derivatives plug into whichever rule applies.
For a specific point, substitute the x-value AFTER differentiating. Use exact values from the unit circle: sin(π/4) = √2/2, sin(π/6) = 1/2, cos(π/3) = 1/2, tan(π/4) = 1, etc.
Linear-argument shortcuts (not in the booklet)
For expressions like sec(ax+b), cosec(ax+b), cot(ax+b), the chain rule just multiplies the formula-booklet derivative by a:
sec — linear argument
ddx[sec(ax+b)] = a sec(ax+b) tan(ax+b)
SAME argument inside both trig factors
cosec — linear argument
ddx[cosec(ax+b)] = −a cosec(ax+b) cot(ax+b)
SAME argument inside both trig factors
Worked examples
WE 1
Prove d/dx[cosec x] = −cosec x cot x from first definitions
Show that the derivative of y = cosec x is −cosec x · cot x.
Step 1 — rewrite using the definitiony = cosec x = 1/sin xStep 2 — apply the quotient rule. numerator = 1, denominator = sin xd/dx[u/v] = (v · u’ − u · v’)/v²u = 1, u’ = 0; v = sin x, v’ = cos xdy/dx = (sin x · 0 − 1 · cos x)/sin²x = −cos x/sin²xStep 3 — split into the reciprocal trig form−cos x/sin²x = −(1/sin x) · (cos x/sin x) = −cosec x · cot x∴ d/dx[cosec x] = −cosec x cot x ✓the “trick” in the last step is splitting cos x/sin²x as TWO factors: one is 1/sin x (which is cosec x), the other is cos x/sin x (which is cot x). Look for this kind of splitting in any reciprocal trig proof.
WE 2
Chain rule — sec with a linear argument
Find dydx for y = 4 sec(2x − π/6).
Step 1 — base derivative from formula bookletd/dx[sec u] = sec u · tan u (where u = 2x − π/6)Step 2 — chain rule: multiply by du/dxu = 2x − π/6 ⟹ du/dx = 2Step 3 — combine, including the constant 4dy/dx = 4 · sec(2x − π/6) · tan(2x − π/6) · 2 = 8 sec(2x − π/6) tan(2x − π/6)dy/dx = 8 sec(2x − π/6) tan(2x − π/6)notice both sec and tan share the SAME argument 2x − π/6. A common error is to write 8 sec(2x − π/6) tan(x) — the chain rule keeps the inner expression intact in both factors.
WE 3
Product rule with a reciprocal trig factor
Find dydx for y = x sec x.
Step 1 — identify the two factors for product ruleu = x, u’ = 1v = sec x, v’ = sec x · tan xStep 2 — apply product rule: (uv)’ = u’v + uv’dy/dx = (1)(sec x) + (x)(sec x tan x) = sec x + x sec x tan xStep 3 — factor sec x for cleaner form = sec x (1 + x tan x)dy/dx = sec x (1 + x tan x)factor sec x out at the end if asked for a “simplified” form. Otherwise both expressions (factored and expanded) are correct.
WE 4
Chain rule with a NONLINEAR argument
Find dydx for y = cot(3x² + 1).
Step 1 — base derivative from formula bookletd/dx[cot u] = −cosec² u (where u = 3x² + 1)Step 2 — chain rule: multiply by du/dxu = 3x² + 1 ⟹ du/dx = 6xStep 3 — combinedy/dx = −cosec²(3x² + 1) · 6x = −6x cosec²(3x² + 1)dy/dx = −6x cosec²(3x² + 1)the inner derivative du/dx = 6x is no longer a constant — that’s the only thing nonlinear arguments change. Everything else (formula booklet result, the negative sign for cot) is identical to the linear case.
WE 5
Gradient at a specific value — exact form
Find the exact value of dydx for y = cosec x at the point where x = π/4.
Step 1 — derivative from formula bookletdy/dx = −cosec x · cot xStep 2 — exact values at x = π/4sin(π/4) = √2/2 ⟹ cosec(π/4) = 1/(√2/2) = 2/√2 = √2tan(π/4) = 1 ⟹ cot(π/4) = 1Step 3 — substitutedy/dx |x = π/4 = −(√2)(1) = −√2dy/dx = −√2at x = π/4 the gradient is negative — consistent with cosec x decreasing from its asymptote at x = 0 toward its minimum at x = π/2. Keeping the answer as −√2 is “exact form”; the decimal −1.414… would lose marks on a Paper-1 question asking for exact value.
WE 6
Equation of tangent at a specific point
Find the equation of the tangent to the curve y = cot x at the point where x = π/4.
Step 1 — find the y-coordinatey at x = π/4: cot(π/4) = 1 ⟹ point is (π/4, 1)Step 2 — derivativedy/dx = −cosec² xStep 3 — gradient at the pointcosec(π/4) = √2 ⟹ cosec²(π/4) = 2dy/dx |x = π/4 = −2Step 4 — tangent equation y − y₀ = m(x − x₀)y − 1 = −2(x − π/4)y − 1 = −2x + π/2y = −2x + π/2 + 1tangent: y = −2x + π/2 + 1 (or 2x + y = π/2 + 1)keep π/2 as an exact constant — don’t approximate to 1.571. The intercept has both a numerical and a π part, which is normal for trig-based tangent equations.
💡 Top tips
Always reach for the formula booklet — even if you “know” the derivatives. The minus signs and squared terms are easy to confuse under exam stress.
Two-trig-function pattern: sec’ and cosec’ are products of TWO trig functions (sec·tan and cosec·cot); cot’ is a single squared function (cosec²). Recognise which one you’re using.
Same argument inside both factors when using chain rule: d/dx[sec(3x)] = 3 sec(3x) tan(3x) — the 3x stays inside tan, not just inside sec.
Use exact values from the unit circle at common angles π/6, π/4, π/3, π/2, etc. Avoid decimal approximations on Paper-1 questions.
When combining with product or quotient rule, do the reciprocal-trig derivative cleanly first, then plug it into the rule. Don’t try to do everything at once.
⚠ Common mistakes
Forgetting the negative sign on cosec or cot derivatives. Both “co-” functions need a minus; sec does NOT.
Confusing sec and cosec derivatives: sec → sec·tan; cosec → −cosec·cot. The PAIRS are: (sec, tan) and (cosec, cot). Don’t mix.
Writing cosec² x as cosec x² or (cosec x)² — these are different. cosec² x means (cosec x)², not cosec(x²).
Chain-rule arguments mixed up: d/dx[cosec 3x] = −3 cosec 3x cot 3x, not −3 cosec x cot 3x. Both trig factors must carry the same inner argument.
Forgetting the inner derivative on nonlinear arguments — d/dx[cot(x²)] = −cosec²(x²) · 2x, NOT just −cosec²(x²). The 2x factor is essential.
Up next: Differentiating Inverse Trigonometric Functions. arcsin, arccos, arctan are the inverses of sine, cosine, tangent — and their derivatives are surprisingly algebraic, not trigonometric. d/dx[arcsin x] = 1/√(1−x²), d/dx[arccos x] = −1/√(1−x²), d/dx[arctan x] = 1/(1+x²). All proved using the inverse-function technique you saw two notes ago.
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