01 · A scaled angle
Differentiate 4tan(3x − 1).
Hint
Include the outside coefficient 4 and the inner derivative 3.
Worked solution
The derivative is 12sec²(3x − 1), where cos(3x − 1) ≠ 0.
Understand · explore · practise
Differentiating tan, sec, cosec and cot
Derive and use the derivatives of tan, sec, cosec and cot. Explore one branch at a time, combine rules for composite expressions and practise exact tangents and stationary points.
Before you startReciprocal trigonometric functions; radians; chain, product and quotient rules
01 / Compare four derivatives
The formula’s sign and the function’s value are different things.
d/dx(tan x) = sec²x
d/dx(sec x) = sec x tan x
d/dx(cosec x) = −cosec x cot x
d/dx(cot x) = −cosec²x
All angles are in radians and each formula applies only where the original function exists.
The derivatives of tan and cot have fixed signs on each valid branch: tan increases and cot decreases. The derivatives of sec and cosec can change sign because they contain unsquared factors. Explore their values and slopes separately.
Function height and derivative heightExplore
d/dx(tan x) = sec²x.
x = 0π/12; f(x) = 0; f′(x) = 1.
The graph is increasing at the selected point because its derivative is positive.
Angles are in radians. Each window lies inside one valid branch; it does not reach a pole or join across one. The green height is the gold tangent’s gradient, not the original function’s height.
02 / Derive the tangent derivative
Write tan x as sin x divided by cos x.
Use the quotient ruleWorked example
d/dx(tan x) = [cos²x + sin²x]/cos²x
Differentiating cosine creates a second minus sign.
d/dx(tan x) = 1/cos²x = sec²x
Use the Pythagorean identity.
Exclude x = π/2 + kπ, k an integer
These are the poles of the original tangent function.
03 / Check the secant sign
Differentiating 1/cos x produces a positive formula.
Use a negative powerWorked example
sec x = (cos x)⁻¹
The original denominator must be non-zero.
d/dx(sec x) = −(cos x)⁻²(−sin x)
The outside and inside derivatives both contribute a minus sign.
d/dx(sec x) = sin x/cos²x = sec x tan x
There is no leading minus sign in the final formula.
This does not mean sec x is increasing everywhere. For −π/2 < x < 0, sec x is positive while tan x is negative, so its derivative is negative.
Watch secant’s slope change across a branch
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Secant with a linear angle
Differentiate 5sec(2x + 1).
Hint
Keep the positive sec·tan formula and multiply by 2.
Worked solution
The derivative is 10sec(2x + 1)tan(2x + 1), where cos(2x + 1) ≠ 0.
03 · Height versus slope
For y = sec x at x = −π/4, find the value and gradient.
Hint
sec(−π/4) = √2 and tan(−π/4) = −1.
Worked solution
The value is √2 and the gradient is −√2. A positive height can have a negative slope.
04 / Differentiate cosecant
The sine derivative leaves one minus sign.
cosec x = (sin x)⁻¹Worked example
d/dx(cosec x) = −cos x/sin²x
Use the chain rule.
d/dx(cosec x) = −cosec x cot x
Separate the reciprocal sine and cotangent factors.
Exclude x = kπ, k an integer
Sine is zero at these angles.
04 · A negative angle factor
Differentiate cosec(1 − 4x).
Hint
The formula contributes −1 and the inner derivative contributes −4.
Worked solution
The derivative is 4cosec(1 − 4x)cot(1 − 4x), where sin(1 − 4x) ≠ 0.
05 / Differentiate cotangent
Its derivative is always negative on its domain.
cot x = cos x/sin xWorked example
d/dx(cot x) = [−sin²x − cos²x]/sin²x
Use the quotient rule in this order.
d/dx(cot x) = −1/sin²x = −cosec²x
The squared cosecant is positive.
Exclude x = kπ
Cotangent decreases on each interval between adjacent poles.
05 · Cotangent derivative
Differentiate 2cot(5x).
Hint
Use both the inner factor 5 and the negative cotangent rule.
Worked solution
The derivative is −10cosec²(5x), where sin(5x) ≠ 0.
06 · Local direction
Is cot x increasing or decreasing at x = 2π/3? Find the exact gradient.
Hint
sin²(2π/3) = 3/4.
Worked solution
The gradient is −4/3, so cot x is decreasing there.
06 / Differentiate nonlinear angles and powers
A square outside a trig function adds another layer.
d/dx[tan g] = sec²(g)g′
d/dx[sec g] = sec(g)tan(g)g′
d/dx[cosec g] = −cosec(g)cot(g)g′
d/dx[cot g] = −cosec²(g)g′
Differentiate sec²(3x)Worked example
y = [sec(3x)]²
The square is outside the secant function.
y′ = 2sec(3x) × sec(3x)tan(3x) × 3
Differentiate all three layers.
y′ = 6sec²(3x)tan(3x)
Keep cos(3x) ≠ 0.
07 · A nonlinear angle
Differentiate tan(x² + 1).
Hint
The inside derivative is 2x.
Worked solution
The derivative is 2x sec²(x² + 1), where cos(x² + 1) ≠ 0.
08 · A power of cosecant
Differentiate cosec³(2x).
Hint
Differentiate the cube, then cosecant, then 2x.
Worked solution
The derivative is −6cosec³(2x)cot(2x), where sin(2x) ≠ 0.
07 / Use product rule when factors multiply
Choose the outside operation first.
Differentiate x sec(2x)Worked example
u′ = 1; v′ = 2sec(2x)tan(2x)
The second factor needs the chain rule.
y′ = sec(2x) + 2x sec(2x)tan(2x)
Use both product terms.
y′ = sec(2x)[1 + 2x tan(2x)]
The original domain requires cos(2x) ≠ 0.
09 · A cotangent product
Differentiate x²cot x.
Hint
The cotangent derivative is negative.
Worked solution
y′ = 2x cot x − x²cosec²x, where sin x ≠ 0.
10 · Two trig factors
Differentiate sec x tan x and simplify the answer.
Hint
Use product rule, then factor sec x.
Worked solution
y′ = sec x tan²x + sec³x = sec x(tan²x + sec²x), where cos x ≠ 0. This is also the second derivative of sec x.
08 / Simplify while retaining the original domain
Equivalent expressions can have different natural domains.
Differentiate tan x/sec xWorked example
tan x/sec x = sin x where cos x ≠ 0
Both original functions must exist before simplification.
y′ = cos x on the original domain
The simplified expression is easy to differentiate.
The original quotient remains undefined at π/2 + kπ
The fact that sine is defined there does not fill those holes.
11 · A quotient with holes
Differentiate cot x/cosec x on its original domain.
Hint
The quotient simplifies to cos x only where sin x ≠ 0.
Worked solution
The derivative is −sin x for x ≠ kπ. The original quotient remains undefined at integer multiples of π.
12 · An identity shortcut
Differentiate sec²x − tan²x and state the restriction.
Hint
This expression is identically 1 wherever both terms exist.
Worked solution
The derivative is 0 where cos x ≠ 0. Direct differentiation gives 2sec²x tan x − 2tan x sec²x = 0.
09 / Find exact tangents and normals
Check the angle is in the original domain.
Tangent to y = sec x at x = π/3Worked example
The point is (π/3, 2)
Use cos(π/3) = 1/2.
y′ = sec x tan x, so m = 2√3
Evaluate both factors.
y − 2 = 2√3(x − π/3)
Keep the exact angle and gradient.
13 · A cosecant normal
Find the normal to y = cosec x at x = π/6.
Hint
The point is (π/6, 2), and the tangent gradient is −2√3.
Worked solution
The normal has gradient 1/(2√3). Its equation is y − 2 = (x − π/6)/(2√3).
14 · A cotangent tangent
Find the tangent to y = cot(2x) at x = π/8.
Hint
The angle inside cotangent is π/4.
Worked solution
The point is (π/8, 1) and the gradient is −2cosec²(π/4) = −4. The tangent is y − 1 = −4(x − π/8).
10 / Find all stationary points in the interval
Exclude poles before solving a derivative equation.
Stationary points of y = sec x for 0 ≤ x ≤ 2πWorked example
y′ = sec x tan x; sec x is never zero
Stationary inputs require tan x = 0.
x = 0, π, 2π
The poles π/2 and 3π/2 are excluded from the domain.
Points: (0, 1), (π, −1), (2π, 1)
The underlying secant function has minima at 0 and 2π and a maximum at π.
If asked for interior stationary points of the restricted interval, give only x = π
Always follow the wording about endpoints.
Classify locally on the continuous branch. Never draw a derivative sign interval straight through a pole as though the function were continuous there.
15 · A cosecant stationary point
Find and classify all stationary points of y = cosec x for 0 < x < π.
Hint
cosec x is never zero, so the derivative is zero only when cot x = 0.
Worked solution
The only point is (π/2, 1). Its derivative is negative before π/2 and positive after it, so this is a minimum.
16 · No stationary points
Does y = tan(2x) have any stationary points on its real domain?
Hint
Its derivative is 2sec²(2x).
Worked solution
No. The derivative is strictly positive at every valid input. Points where cos(2x) = 0 are excluded poles, not stationary points.
11 / Check formula, angle and domain
The derivative of sec is positive sec·tan; the product itself may be negative.
- Use radians for the standard formulas.
- For tan and sec, exclude zeros of cosine; for cosec and cot, exclude zeros of sine.
- Multiply by the derivative of any inside angle.
- Apply product or quotient rule to combinations of factors.
- Keep original exclusions after identities or cancellation.
- Use derivative signs within a continuous branch.
Section 1 of 11 · Compare four derivatives