What is stationary points of a trigonometric curve?

To locate stationary points you still solve f'(x) = 0, but now the solutions are angles and there are infinitely many of them. For y = sin x + cos x, y' = cos x - sin x = 0 gives tan x = 1, so x = π4 + nπ for integer n. Within a stated domain such as 0 x 2π you list only the solutions that fall inside, here x = π4 (a maximum) and x = 5π4 (a minimum). Always respect the given domain and remember that periodic functions repeat their stationary points.

TASC TCE 2024-style practice: Calculator-free. Differentiate y = 3cos(π x) + 6tan(2x).

Differentiate term by term, applying the standard derivatives with the chain rule for the inner linear functions. Term 1: ddx[3cos(π x)] = 3×(-sin(π x))× π = -3πsin(π x). The chain rule brings down the factor π, the derivative of the inside π x. Term 2: ddx[6tan(2x)] = 6sec^2(2x)× 2 = 12sec^2(2x). The derivative of tan u is sec^2u, and the chain rule brings down the factor 2. So dydx = -3πsin(π x) + 12sec^2(2x). One mark is for each correctly differentiated term, including the chain-rule factors. These results require x to be in radians.

TASC TCE 2023-style practice: Calculator-free. Differentiate (no simplification required): y = ln(4x + 1) + sin(3x^2).

Differentiate each term using the chain rule. Term 1: ddx[ln(4x + 1)] = 14x + 1× 4 = 44x + 1. The derivative of ln u is u'u with u = 4x + 1, so u' = 4. Term 2: ddx[sin(3x^2)] = cos(3x^2)× 6x = 6xcos(3x^2). The derivative of sin u is cos u× u', with u = 3x^2 giving u' = 6x. So dydx = 44x + 1 + 6xcos(3x^2). One mark per term. Markers accept the unsimplified form here, but the chain-rule factor on each term is essential.

§-Syllabus dot point

TASMath MethodsSyllabus dot point

How do we differentiate sine, cosine and tangent, and combine them with the chain, product and quotient rules?

Establish and use the derivatives of sin x, cos x and tan x, including with the chain, product and quotient rules.

Derivatives of sin x, cos x and tan x, why radians are required, and how to combine them with the chain, product and quotient rules for TCE Mathematics Methods Unit 3.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

This dot point completes your library of standard derivatives by adding the three core trigonometric functions. Once you have these alongside powers, exponentials and logarithms, you can differentiate almost any function in the course.

The three standard derivatives

The results you must memorise are short, but the minus sign and the radian requirement trip up many students.

Notice the symmetry: differentiating sine gives cosine, and differentiating cosine gives negative sine. Differentiating twice more returns you to where you started, so the fourth derivative of $\sin x$ is $\sin x$ again.

Combining with the chain rule

For a function of the form $\sin(f(x))$ , the chain rule multiplies by the derivative of the inside:

\frac{d}{dx}\big(\sin(f(x))\big) = f'(x)\cos(f(x)), \qquad \frac{d}{dx}\big(\cos(f(x))\big) = -f'(x)\sin(f(x)).

So $\dfrac{d}{dx}\sin(3x) = 3\cos(3x)$ and $\dfrac{d}{dx}\cos(x^2) = -2x\sin(x^2)$ . The inner derivative is exactly the factor students most often drop.

Combining with product and quotient rules

Trigonometric terms appear inside products and quotients constantly. Apply the product rule $u'v + uv'$ or the quotient rule $\dfrac{u'v - uv'}{v^2}$ exactly as for any other functions, treating $\sin x$ , $\cos x$ or $\tan x$ as one of the factors.

Differentiating a product and a quotient

Differentiate (a) $y = x^2\cos x$ and (b) $y = \dfrac{\sin x}{x}$ .

Step 1: Identify the rule needed for each expression

Part (a) is a product of two functions, so the product rule $\dfrac{d}{dx}(uv) = u'v + uv'$ applies. Part (b) is a quotient, so the quotient rule $\dfrac{d}{dx}\!\left(\dfrac{u}{v}\right) = \dfrac{u'v - uv'}{v^2}$ applies. Naming $u$ and $v$ before differentiating keeps the working organised.

Step 2: Differentiate the product $y = x^2\cos x$

Set $u = x^2$ and $v = \cos x$ , giving $u' = 2x$ and $v' = -\sin x$ (remember the minus sign on the derivative of cosine). Applying the product rule:

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

Step 3: Differentiate the quotient $y = \dfrac{\sin x}{x}$

Set $u = \sin x$ and $v = x$ , giving $u' = \cos x$ and $v' = 1$ . The quotient rule puts $u'v - uv'$ in the numerator and $v^2$ in the denominator:

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

Final answers: (a) $\dfrac{dy}{dx} = 2x\cos x - x^2\sin x$ ; (b) $\dfrac{dy}{dx} = \dfrac{x\cos x - \sin x}{x^2}$ .

Higher derivatives and the cycle of four

Because differentiating sine and cosine cycles through four functions, the higher derivatives of $\sin x$ follow a repeating pattern of period four:

\frac{d}{dx}\sin x = \cos x, \quad \frac{d^{2}}{dx^{2}}\sin x = -\sin x, \quad \frac{d^{3}}{dx^{3}}\sin x = -\cos x, \quad \frac{d^{4}}{dx^{4}}\sin x = \sin x.

This is why

y = \sin x

satisfies

y'' = -y

, the differential equation of simple harmonic motion. Any oscillating model built from sine and cosine inherits this self-replicating structure, which is what makes these functions the natural language for waves, vibrations and tides.

Stationary points of a trigonometric curve

To locate stationary points you still solve $f'(x) = 0$ , but now the solutions are angles and there are infinitely many of them. For $y = \sin x + \cos x$ , $y' = \cos x - \sin x = 0$ gives $\tan x = 1$ , so $x = \dfrac{\pi}{4} + n\pi$ for integer $n$ . Within a stated domain such as $0 \le x \le 2\pi$ you list only the solutions that fall inside, here $x = \dfrac{\pi}{4}$ (a maximum) and $x = \dfrac{5\pi}{4}$ (a minimum). Always respect the given domain and remember that periodic functions repeat their stationary points.

Where these are used

The derivatives of trigonometric functions feed straight into curve sketching (finding stationary points of $y = \sin x + \cos x$ , for example), into optimisation problems with periodic models, and into rates of change for anything oscillating such as a tide height or an alternating current. They also pair with integration in Unit 4, where reversing each derivative gives the corresponding antiderivative.

Summary

Learn the three standard derivatives, keep your calculator in radian mode, and treat the trigonometric function as just another piece when you apply the chain, product or quotient rule. The two recurring slips are the missing minus sign on the derivative of cosine and the missing inner factor from the chain rule, so build a habit of checking both every time.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20242 marksCalculator-free. Differentiate

y = 3\cos(\pi x) + 6\tan(2x)

Show worked answer →

Differentiate term by term, applying the standard derivatives with the chain rule for the inner linear functions.

Term 1: $\dfrac{d}{dx}\big[3\cos(\pi x)\big] = 3\times(-\sin(\pi x))\times \pi = -3\pi\sin(\pi x)$ . The chain rule brings down the factor $\pi$ , the derivative of the inside $\pi x$ .

Term 2: $\dfrac{d}{dx}\big[6\tan(2x)\big] = 6\sec^{2}(2x)\times 2 = 12\sec^{2}(2x)$ . The derivative of $\tan u$ is $\sec^{2}u$ , and the chain rule brings down the factor $2$ .

So $\dfrac{dy}{dx} = -3\pi\sin(\pi x) + 12\sec^{2}(2x)$ . One mark is for each correctly differentiated term, including the chain-rule factors. These results require $x$ to be in radians.

TCE 20232 marksCalculator-free. Differentiate (no simplification required):

y = \ln(4x + 1) + \sin(3x^{2})

Show worked answer →

Differentiate each term using the chain rule.

Term 1: $\dfrac{d}{dx}\big[\ln(4x + 1)\big] = \dfrac{1}{4x + 1}\times 4 = \dfrac{4}{4x + 1}$ . The derivative of $\ln u$ is $\dfrac{u'}{u}$ with $u = 4x + 1$ , so $u' = 4$ .

Term 2: $\dfrac{d}{dx}\big[\sin(3x^{2})\big] = \cos(3x^{2})\times 6x = 6x\cos(3x^{2})$ . The derivative of $\sin u$ is $\cos u\times u'$ , with $u = 3x^{2}$ giving $u' = 6x$ .

So $\dfrac{dy}{dx} = \dfrac{4}{4x + 1} + 6x\cos(3x^{2})$ . One mark per term. Markers accept the unsimplified form here, but the chain-rule factor on each term is essential.

TCE 20224 marksCalculator-assumed. A tide height (in metres) is modelled by

h(t) = 2.5 + 1.8\sin\!\left(\dfrac{\pi t}{6}\right)

, where

t

is the time in hours after midnight. (a) Find

h'(t)

. (b) Find the rate at which the tide is rising or falling at

t = 2

, to two decimal places, and state whether it is rising or falling.

Show worked answer →

(a) Differentiate using the chain rule on the inner function $\dfrac{\pi t}{6}$ , whose derivative is $\dfrac{\pi}{6}$ :

h'(t) = 1.8\cos\!\left(\frac{\pi t}{6}\right)\times \frac{\pi}{6} = 0.3\pi\cos\!\left(\frac{\pi t}{6}\right).

(b) Substitute $t = 2$ , so $\dfrac{\pi t}{6} = \dfrac{\pi}{3}$ and $\cos\dfrac{\pi}{3} = 0.5$ :

h'(2) = 0.3\pi\times 0.5 = 0.15\pi \approx 0.47 \text{ m/h}.

The rate is positive, so the tide is rising at about

0.47

metres per hour. Markers reward the chain-rule factor

\dfrac{\pi}{6}

, the correct derivative, and the signed interpretation with units.