Derivatives of trigonometric functions in radians (QCE Mathematical Methods Unit 3)

Syllabus dot point

“Differentiate trigonometric functions, including compositions of the form $\sin(f(x))$ , $\cos(f(x))$ and $\tan(f(x))$ , working in radians”

7 min readUpdated 7 Oct 2026

Quick answer

In radians, $\sin x$ , $\cos x$ and $\tan x$ differentiate to $\cos x$ , $-\sin x$ and $\sec^2 x$ respectively, and the chain rule extends this to $\frac{d}{dx} \sin(f(x)) = f'(x) \cos(f(x))$ and analogous forms for $\cos$ and $\tan$ .

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What this dot point is asking
The answer
Exam-style questions

What this dot point is asking

QCAA wants you to differentiate trigonometric functions and their compositions in radians, including combinations with the product, quotient and chain rules. Trigonometric derivatives appear in Paper 1 short answer, in Paper 2 modelling questions about oscillating quantities (tides, sound, AC currents, planetary orbits), and in PSMTs that involve periodic phenomena.

The answer

Why radians

In Methods, all calculus on trig functions is done with $x$ in radians. The formula $\frac{d}{dx}(\sin x) = \cos x$ is only true when $x$ is in radians. If you work in degrees the derivative picks up an awkward factor of $\frac{\pi}{180}$ , which is why QCAA requires radians for calculus questions and why most calculators default to degree mode on a fresh reset (always check and switch to radians).

The standard derivatives

\frac{d}{dx}(\sin x) = \cos x

\frac{d}{dx}(\cos x) = -\sin x

\frac{d}{dx}(\tan x) = \sec^2 x = \frac{1}{\cos^2 x}

The minus sign on the derivative of $\cos$ is the single most common Paper 1 trap. Memorise it.

Chain-rule generalisations

For any differentiable inner $f(x)$ :

\frac{d}{dx}\bigl(\sin(f(x))\bigr) = f'(x) \cos(f(x))

\frac{d}{dx}\bigl(\cos(f(x))\bigr) = -f'(x) \sin(f(x))

\frac{d}{dx}\bigl(\tan(f(x))\bigr) = f'(x) \sec^2(f(x))

The most common case is a linear inner function, where the chain rule factor is just a constant.

Linear inside argument

For constants $a$ and $b$ :

\frac{d}{dx}\bigl(\sin(a x + b)\bigr) = a \cos(a x + b)

\frac{d}{dx}\bigl(\cos(a x + b)\bigr) = -a \sin(a x + b)

This is the most heavily examined form. Almost every modelling question is of the type $A \sin(\omega t + \phi) + C$ .

Where the standard derivatives come from

The derivative of $\sin x$ being $\cos x$ follows from first principles using the limit $\lim_{h \to 0}\dfrac{\sin h}{h} = 1$ , which holds only when $h$ is in radians. That single limit is the reason calculus on trigonometric functions must use radians: in degrees the limit is $\dfrac{\pi}{180}$ instead of $1$ , and the clean derivative formulas break. Knowing this justifies the radian requirement rather than treating it as an arbitrary rule.

Velocity, acceleration and oscillation

Differentiating a sinusoidal displacement gives a sinusoidal velocity, and differentiating again gives acceleration. For $s(t) = A\sin(\omega t)$ , the velocity is $A\omega\cos(\omega t)$ and the acceleration is $-A\omega^2\sin(\omega t) = -\omega^2 s$ . The acceleration being proportional to the negative of the displacement is the defining equation of simple harmonic motion, which is why sinusoidal models describe tides, sound and oscillating springs. QCAA modelling questions frequently ask for the velocity or the times of maximum speed, both of which come straight from these derivatives.

Worked example

Five differentiation tasks on trigonometric functions in radians, illustrating the chain rule, the product rule, and an application to velocity.

Step 1: Apply the chain rule directly to $y = \cos(3x - 1)$

The outer function is $\cos$ and the inner function is $3x - 1$ . The derivative of $\cos$ is $-\sin$ , and the inner function contributes a factor of $3$ . Notice the minus sign appears because $\frac{d}{dx} \cos = -\sin$ .

\frac{dy}{dx} = -3 \sin(3x - 1)

Step 2: Differentiate when the inner function is a polynomial, $y = \sin(x^2)$

The outer function is $\sin$ and the inner function is $x^2$ . Identifying $f(x) = x^2$ with $f'(x) = 2x$ , the chain rule gives the factor $2x$ in front.

\frac{dy}{dx} = 2x \cos(x^2)

Step 3: Combine the chain and product rules for $y = x^2 \sin x$

The outermost structure is a product of $u = x^2$ and $v = \sin x$ , so we apply the product rule. The inner function of $\sin x$ is simply $x$ , so no chain rule is needed on the trig factor.

\frac{dy}{dx} = u'v + uv' = 2x \sin x + x^2 \cos x

Step 4: Find velocity from a sinusoidal displacement

A particle moves with displacement $s(t) = 4 \sin(2t)$ metres. Velocity is the derivative of displacement, and the chain rule on $\sin(2t)$ produces a factor of $2$ .

v(t) = s'(t) = 4 \times 2 \cos(2t) = 8 \cos(2t)

Substituting $t = \dfrac{\pi}{6}$ and using the exact value $\cos\!\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}$ :

v\!\left(\frac{\pi}{6}\right) = 8 \cos\!\left(\frac{\pi}{3}\right) = 8 \times \frac{1}{2} = 4 \text{ m/s}

Final answer: the particle's velocity at $t = \dfrac{\pi}{6}$ s is $4$ m/s. The exact value $\cos(\pi/3) = \tfrac{1}{2}$ is expected without a calculator on Paper 1.

Step 5: Differentiate $y = \tan(3x)$

The derivative of $\tan$ is $\sec^2$ , and the inner function $3x$ contributes a chain rule factor of $3$ .

\frac{dy}{dx} = 3 \sec^2(3x) = \frac{3}{\cos^2(3x)}

Common traps

Working in degrees: $\frac{d}{dx}(\sin x) = \cos x$ only when $x$ is in radians. Set your calculator to radians for Paper 2. Paper 1 problems are always in radians by convention.
Dropping the minus sign on $\cos$: $\frac{d}{dx}(\cos x) = -\sin x$ . Reversing the sign turns a maximum into a minimum and a velocity into its negative.
Missing the chain rule factor: $\frac{d}{dx}(\sin(2x)) = 2 \cos(2x)$ , not $\cos(2x)$ . The factor of $2$ comes from differentiating the inner $2x$ .
Forgetting exact values: Paper 1 expects $\sin(\pi/6) = 1/2$ , $\cos(\pi/4) = \sqrt{2}/2$ , $\tan(\pi/3) = \sqrt{3}$ and the related angle values without a calculator. Drill the unit circle.
Combining identities incorrectly: $\sin^2 x + \cos^2 x = 1$ , not $\sin^2 x - \cos^2 x = 1$ . Sign errors here lead to wrong derivatives on optimisation questions.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice3 marks

Paper 1 (technique). Differentiate

y = \sin(4x) + \cos(2x)

with respect to

x

Show worked answer →

Differentiate term by term using the chain rule.

For $\sin(4x)$ : inner function $4x$ has derivative $4$ , so $\frac{d}{dx} \sin(4x) = 4 \cos(4x)$ .

For $\cos(2x)$ : inner function $2x$ has derivative $2$ , so $\frac{d}{dx} \cos(2x) = -2 \sin(2x)$ .

$\frac{dy}{dx} = 4 \cos(4x) - 2 \sin(2x).$

Markers reward both chain rule factors and the correct minus sign on the derivative of $\cos$ . Forgetting either factor of $2$ or $4$ is the most common Paper 1 mistake on this style of question.

Practice4 marks

Paper 1 (technique). Given

f(x) = \sin(2x) \cos(2x)

, (a) determine

f'(x)

using the product rule and (b) verify by first rewriting

f(x)

using the double-angle identity.

Show worked answer →

(a) Product rule with $u = \sin(2x)$ , $v = \cos(2x)$ , $u' = 2 \cos(2x)$ , $v' = -2 \sin(2x)$ .

$f'(x) = u' v + u v' = 2 \cos(2x) \cdot \cos(2x) + \sin(2x) \cdot (-2 \sin(2x))$
$= 2 \cos^2(2x) - 2 \sin^2(2x) = 2 \cos(4x).$

(b) Identity: $\sin(2x) \cos(2x) = \frac{1}{2} \sin(4x)$ . Differentiate: $f'(x) = \frac{1}{2} \cdot 4 \cos(4x) = 2 \cos(4x)$ . Matches.

Markers reward both methods reaching the same answer, with explicit use of $\cos^2 - \sin^2 = \cos(2 \cdot \text{angle})$ in part (a). The verification in part (b) shows mathematical maturity and earns the final mark.

Practise this

Sources & how we know this

Mathematical Methods General senior syllabus 2025 — QCAA

What this dot point is asking

The answer

Why radians

The standard derivatives

Chain-rule generalisations

Linear inside argument

Where the standard derivatives come from

Velocity, acceleration and oscillation

Step 1: Apply the chain rule directly to y=cos⁡(3x−1)y = \cos(3x - 1)y=cos(3x−1)

Step 2: Differentiate when the inner function is a polynomial, y=sin⁡(x2)y = \sin(x^2)y=sin(x2)

Step 3: Combine the chain and product rules for y=x2sin⁡xy = x^2 \sin xy=x2sinx

Step 4: Find velocity from a sinusoidal displacement

Step 5: Differentiate y=tan⁡(3x)y = \tan(3x)y=tan(3x)

Exam-style questions

Practise this

Sources & how we know this

Step 1: Apply the chain rule directly to $y = \cos(3x - 1)$

Step 2: Differentiate when the inner function is a polynomial, $y = \sin(x^2)$

Step 3: Combine the chain and product rules for $y = x^2 \sin x$

Step 5: Differentiate $y = \tan(3x)$