Unit 3 · Topic 2.1 · Differentiation of trigonometric functions and differentiation rules

Trigonometric functions

Trig is guaranteed to show up in your external. This guide starts with the Unit 1 and 2 trig you need to be fluent with, then moves on to differentiating sine and cosine and using them in models. Tick off each skill once you can do it without help.

Skills in this topic

Calculus of trigonometric functions, about 5 hours of class time. Only tick a skill when you could do it in an exam.

Use the rules $\frac{d}{dx}\sin x=\cos x$ and $\frac{d}{dx}\sin f(x)=f'(x)\cos f(x)$. Part 2

Use the rules $\frac{d}{dx}\cos x=-\sin x$ and $\frac{d}{dx}\cos f(x)=-f'(x)\sin f(x)$. Part 2

Model and solve problems that involve derivatives of trigonometric functions, with and without technology. Part 2

Part 1

The Unit 1 and 2 trig you need

Anything from Units 1 and 2 can appear in the external, but trig is the one you can count on. Before you differentiate anything, make sure you are fluent with the unit circle, exact values, solving trig equations and building trig models.

Watch it first

The sine curve is the height of a point on the unit circle

As the point travels round the circle, its height is $\sin\theta$ and its horizontal position is $\cos\theta$. One full lap is $2\pi$, which is why the period of $\sin\theta$ is $2\pi$.

The unit circle

You will refer to the unit circle constantly when you model with trig. The point at angle $\theta$ is $(\cos\theta,\sin\theta)$, so:

$x=\cos\theta$$y=\sin\theta$$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$

Know which quadrants make sine and cosine positive, and what the related angle is in each one. It flows into everything else in this part.

Special triangles and exact values

Finding exact values is one of the Year 11 skills you will use most in the external. Know your two special triangles. If you would rather not draw them out every time, learn the table as well.

$1:\sqrt{3}:2$

$1:1:\sqrt{2}$

Degrees0°30°45°60°90°180°360°

Radians$0$$\frac{\pi}{6}$$\frac{\pi}{4}$$\frac{\pi}{3}$$\frac{\pi}{2}$$\pi$$2\pi$

$\sin$$0$$\frac{1}{2}$$\frac{1}{\sqrt2}$$\frac{\sqrt3}{2}$$1$$0$$0$

$\cos$$1$$\frac{\sqrt3}{2}$$\frac{1}{\sqrt2}$$\frac{1}{2}$$0$$-1$$1$

$\tan$$0$$\frac{1}{\sqrt3}$$1$$\sqrt3$undefined$0$$0$

Solving trig equations

When you model with trig, you will often need to solve a trig equation along the way. Work through these steps every time.

1Isolate the trig function on one side of the equation.
2Find the base angle using the special triangles.
3Use the unit circle to work out which quadrants the solutions are in.
4Account for any change to the domain, like $3x$ or $2x-\frac{\pi}{4}$ inside the function.
5Check that every solution lies inside the given domain.

Try these before you open the solutions

Practice 1: exact values

Technology-free

Solve each equation to get exact values for $x$.

(a) $\sin(x)=\dfrac{\sqrt3}{2},\ 0\le x\le 2\pi$ (b) $\sqrt2\cos(x)+1=0,\ 0\le x\le 2\pi$ (c) $\sqrt3-3\tan(x)=0,\ -2\pi\le x\le 2\pi$

(a) The base angle is $\frac{\pi}{3}$. Sine is positive in quadrants 1 and 2, so $x=\frac{\pi}{3}$ and $x=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$.

(b) Isolate first: $\cos(x)=-\frac{1}{\sqrt2}$. The base angle is $\frac{\pi}{4}$. Cosine is negative in quadrants 2 and 3, so $x=\pi-\frac{\pi}{4}$ and $x=\pi+\frac{\pi}{4}$.

(c) Isolate: $\tan(x)=\frac{\sqrt3}{3}=\frac{1}{\sqrt3}$. The base angle is $\frac{\pi}{6}$, and tan is positive in quadrants 1 and 3. That gives $\frac{\pi}{6}$ and $\frac{7\pi}{6}$, and the domain goes back to $-2\pi$, so subtract $2\pi$ from each as well.

Answer

(a) $\frac{\pi}{3},\frac{2\pi}{3}$ · (b) $\frac{3\pi}{4},\frac{5\pi}{4}$ · (c) $-\frac{11\pi}{6},-\frac{5\pi}{6},\frac{\pi}{6},\frac{7\pi}{6}$

Practice 2: a changed domain

Technology-free

(a) Solve $\cos(3x)=-\dfrac{1}{2}$ for $0\le x\le 2\pi$. (b) Use substitution to solve $\tan\left(2x-\dfrac{\pi}{4}\right)=-1$ for $0\le x\le\pi$.

(a) Change the domain first. If $0\le x\le 2\pi$, then $0\le 3x\le 6\pi$, which is three laps of the circle. The base angle is $\frac{\pi}{3}$, and cosine is negative in quadrants 2 and 3:

$$3x=\frac{2\pi}{3},\ \frac{4\pi}{3},\ \frac{8\pi}{3},\ \frac{10\pi}{3},\ \frac{14\pi}{3},\ \frac{16\pi}{3}$$

Divide every one by 3.

(b) Let $\theta=2x-\frac{\pi}{4}$. Then $-\frac{\pi}{4}\le\theta\le\frac{7\pi}{4}$. Tan is $-1$ at $\theta=-\frac{\pi}{4},\ \frac{3\pi}{4},\ \frac{7\pi}{4}$. Now undo the substitution: $2x=\theta+\frac{\pi}{4}=0,\ \pi,\ 2\pi$.

Answer

(a) $\frac{2\pi}{9},\frac{4\pi}{9},\frac{8\pi}{9},\frac{10\pi}{9},\frac{14\pi}{9},\frac{16\pi}{9}$ · (b) $0,\frac{\pi}{2},\pi$

Building trig models

The QCAA often ask you to build your own trig model from a worded situation. For that you need the general form:

$$y=a\cos\big(b(x-h)\big)+k$$

$a$ is the amplitude.

$k$ is the midline, or vertical shift.

$b$ gives the period: $P=\dfrac{2\pi}{b}$.

$h$ is the horizontal shift. You can usually avoid it by choosing the right function.

So how do you choose? Look at where the model starts. If it starts at a maximum or minimum, cosine fits. If it starts on the midline, sine fits. You do not need to memorise which sign goes with which. Picture the unit circle: a negative sine or cosine just goes round the other way.

Try it yourself

Pick where the model starts

Choose a starting point and watch the model pick its function. Then set the amplitude, midline and period to see where $a$, $b$ and $k$ come from.

The model starts

Amplitude Midline

Period (hours)

From the period

In short, when you build a trig model

Look for the amplitude and midline. They are often implied by a maximum and a minimum.
Look for the period, then solve $P=\frac{2\pi}{b}$ for $b$.
Choose the function from where the model starts.

Part 2

Differentiating trigonometric functions

These four rules are fundamental. You need to be completely fluent with them by the time you sit the external.

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

Why the derivative of sine is cosine

Watch the chain rule versions

Same warning as with exponentials and logs: being able to differentiate in this form is not enough on its own. Methods questions build through four phases, and you want to be confident at every one. Pick a phase to see the real questions that match it.

Phase 1 · Apply the rules

Straight use of the rules. Watch the minus sign on the derivative of cosine, because it is the slip I see most often.

For the written-answer questions, the solutions show where each mark goes, based on the QCAA marking guides. Many marks allow follow-through: if an earlier step is wrong but you use your wrong answer correctly, you can still get the later marks. That is another reason to show every step.

2023 Paper 2, Question 1

Multiple choice · Technology-active · [1 mark]

If $f(x)=\sin(3x)$, determine the value of $f'\left(\dfrac{\pi}{8}\right)$.

The 3 comes out the front: $f'(x)=3\cos(3x)$. Then substitute, with your calculator in radians: $3\cos\frac{3\pi}{8}\approx 1.148$.

Answer

(B) $1.148$

2021 Paper 2, Question 8

Multiple choice · Technology-active · [1 mark]

The displacement (in metres) of a particle is given by $s(t)=-3\cos(t)+2\sin(t)$, where $t$ is in seconds. The instantaneous velocity of the particle at time $t=\dfrac{\pi}{2}$ seconds is

Velocity is the derivative of displacement. The two minus signs on the cosine cancel:

$$v(t)=3\sin(t)+2\cos(t),\qquad v\left(\tfrac{\pi}{2}\right)=3(1)+2(0)=3$$

Answer

(D) $3\ \text{m\,s}^{-1}$

2021 Paper 2, Question 10

Multiple choice · Technology-active · [1 mark]

An object travels in a straight line so that its velocity at time $t$ seconds is given by $v(t)=2t+\sin(2t)$. Determine the expression for acceleration as a function of time.

Acceleration is the derivative of velocity. (C) and (D) integrate instead, which would give displacement.

$$a(t)=v'(t)=2+2\cos(2t)$$

Answer

(A) $a(t)=2+2\cos(2t)$

Phase 2 · Add the chain, product and quotient rules

Strictly speaking, this is Topic 2.2, but trig combined with the product and quotient rules shows up all the time, so it is worth seeing here.

2024 Paper 2, Question 6

Multiple choice · Technology-active · [1 mark]

Determine the derivative of $y=2x\cos(3x)$.

Product rule with $u=2x$ and $v=\cos(3x)$, so $u'=2$ and $v'=-3\sin(3x)$. (C) is what you get if you differentiate each part and multiply, which is not the product rule.

$$\frac{dy}{dx}=2\cos(3x)+2x\left(-3\sin(3x)\right)$$

Answer

(A) $2\cos(3x)-6x\sin(3x)$

2021 Paper 1, Question 11(b)

Short response · Technology-free · [3 marks]

Determine the derivative with respect to $x$ of $y=\dfrac{\sin(x)}{x^{2}}$. Give your answer in simplest form.

Which rule? One function is divided by another, so use the quotient rule with $u=\sin x$ and $v=x^{2}$.

1 markChoosing the quotient rule. Saying you will use the product rule is fine too.

$$\frac{dy}{dx}=\frac{x^{2}\cos x-2x\sin x}{\left(x^{2}\right)^{2}}$$

1 markThe derivative.

"Simplest form" means cancelling the common $x$ from the top and bottom.

$$\frac{dy}{dx}=\frac{x\cos x-2\sin x}{x^{3}}$$

1 markSimplest form. Splitting it into $\frac{\cos x}{x^{2}}-\frac{2\sin x}{x^{3}}$ also counts. If an earlier answer was wrong, you can still get this mark for using it correctly.

Answer

$\dfrac{dy}{dx}=\dfrac{x\cos x-2\sin x}{x^{3}}$

2020 Paper 1, Question 11(a)

Short response · Technology-free · [1 mark]

Determine the derivative with respect to $x$ of $y=\dfrac{1}{\sin(x)}$.

Rewrite it as a power so you can use the chain rule: $y=(\sin x)^{-1}$.

$$\frac{dy}{dx}=-1(\sin x)^{-2}\times\cos x$$

1 markThe derivative. Leaving it as $-\cos x\,(\sin x)^{-2}$ is fine.

Answer

$\dfrac{dy}{dx}=-\dfrac{\cos x}{\sin^{2}x}$

Phase 3 · Model with a known equation

These are worded, but the equation is given. Most of the work is turning the time in the question into the right value of $t$, and keeping your calculator in radians.

2025 Paper 2, Question 10

Multiple choice · Technology-active · [1 mark]

The outside air temperature, $T$ (°C), on a particular day at a certain location is modelled by the function $T=25-7\cos\left(\dfrac{\pi t}{12}\right),\ 0\le t\le 24$, where $t$ is the time (hours) since 6:00 am. The rate of change of temperature at 11:30 am is

$t$ is measured from 6:00 am, so 11:30 am is $t=5.5$. It is not $t=11.5$.

$$\frac{dT}{dt}=\frac{7\pi}{12}\sin\left(\frac{\pi t}{12}\right),\qquad T'(5.5)\approx 1.817$$

(A) is $\frac{7\pi}{12}$ on its own, the fastest the temperature ever changes.

Answer

(B) $1.817$ °C per hour

2025 Paper 2, Question 14

Short response · Technology-active · [6 marks]

The number of tourists visiting a country at any given time is modelled by $N(t)=18000\sin\left(\dfrac{\pi}{6}t+6\right)+22000,\ 0\le t\le 12$, where $t$ is the time (months) from the start of the year.

(a) Determine when the number of tourists first reaches 36 000. Provide your response as a decimal number.[1 mark]

(b) Determine the equation of the second derivative of $N(t)$.[2 marks]

(d) Use your result from (c) to explain how the value of the second derivative is consistent with the number of tourists being a minimum.[1 mark]

(a) Solve $N(t)=36000$ on your calculator for $0\le t\le 12$, in radians. The first solution is $t\approx 2.24$ months.

1 markThe time, as a decimal.

(b) Differentiate once. The $\frac{\pi}{6}$ comes out the front:

$$N'(t)=3000\pi\cos\left(\frac{\pi}{6}t+6\right)$$

1 markThe first derivative. Decimals like $9424.8$ are fine.

Differentiate again. Another $\frac{\pi}{6}$ comes out, and $3000\pi\times\frac{\pi}{6}=500\pi^{2}$:

$$N''(t)=-500\pi^{2}\sin\left(\frac{\pi}{6}t+6\right)$$

1 markThe second derivative. If an earlier answer was wrong, you can still get this mark for using it correctly.

1 markThe time of the minimum.

At the minimum the sine part equals $-1$, so $N''=-500\pi^{2}(-1)=500\pi^{2}\approx 4935$. Substituting $t=9.54$ on your calculator gives the same.

1 markThe value of the second derivative, exact or as a decimal. If an earlier answer was wrong, you can still get this mark for using it correctly.

(d) The second derivative is positive, so the graph is concave up there. A turning point where the graph is concave up is a minimum.

1 markExplaining why a positive second derivative fits a minimum.

Answer

(a) $2.24$ months · (b) above · (c) $500\pi^{2}\approx 4935$ · (d) positive, so concave up

2024 Paper 1, Question 9

Multiple choice · Technology-free · [1 mark]

At a certain location, the temperature (°C) can be modelled by the function $T=5\sin\left(\dfrac{\pi}{12}x\right)+23$, where $x$ is the number of hours after sunrise. Determine the rate of change of temperature (°C/hour) when $x=4$.

This is Paper 1, so you need the exact value. Differentiate, then substitute $x=4$:

$$\frac{dT}{dx}=\frac{5\pi}{12}\cos\left(\frac{\pi}{12}x\right),\qquad \frac{5\pi}{12}\cos\frac{\pi}{3}=\frac{5\pi}{12}\times\frac{1}{2}$$

Answer

(B) $\dfrac{5\pi}{24}$ °C per hour

2022 Paper 2, Question 1

Multiple choice · Technology-active · [1 mark]

The position (in cm) of a particle is given by $x=\cos(4t)$, where $t$ is time (in seconds). The velocity of the particle when $t=5$ is

Velocity is the derivative of position, and the 4 comes out the front:

$$v=-4\sin(4t),\qquad v(5)=-4\sin(20)\approx -3.6518$$

Answer

(D) $-3.6518\ \text{cm\,s}^{-1}$

Phase 4 · Model with an unknown equation

This is the one to aim for. Here you often have to build the model yourself, from a graph or a description, before you can differentiate it. Use the three steps from Part 1.

2021 Paper 2, Question 17

Short response · Technology-active · [4 marks]

Rabbits and foxes are among two species of mammals that live on an isolated island. Rabbits represent a significant food source for the foxes. The populations of rabbits and foxes were monitored each month for two years. The graph shows the population of foxes (in thousands) and the population of rabbits (in thousands), at any time $t$ (in months) over the two years. The two populations can be modelled using trigonometric functions.

Foxes Rabbits Population in thousands · time $t$ in months since 1 January

Jane believes that there were periods of time over the two years when the total population of foxes and rabbits on the island exceeded 25 000.

Evaluate the reasonableness of Jane's claim.

Build each model with the three steps. Foxes: maximum 11, minimum 7, so the amplitude is 2 and the midline 9. Max to max is 12 months, so $b=\frac{\pi}{6}$. They start on the midline going up, so use positive sine.

Rabbits: maximum 14.5, minimum 7.5, so the amplitude is 3.5 and the midline 11. The period is also 12 months, and they start at a maximum, so use positive cosine.

$$F(t)=2\sin\left(\frac{\pi}{6}t\right)+9,\qquad R(t)=3.5\cos\left(\frac{\pi}{6}t\right)+11$$

1 markBoth models.

Add them to get the total:

$$T(t)=20+3.5\cos\left(\frac{\pi}{6}t\right)+2\sin\left(\frac{\pi}{6}t\right)$$

1 markThe model for the total. If an earlier answer was wrong, you can still get this mark for using it correctly.

Graph $T(t)$ on your calculator over the two years and find its maximum. It is about 24.03 thousand, at around $t=1$ month, and again a year later.

1 markUsing a suitable representation, like the graph. Solving $T'(t)=0$ for the stationary points also works.

The total never gets to 25 000, so Jane's claim is not reasonable.

1 markDeciding on the claim.

Answer

Not reasonable. The total peaks at about 24 031.

2023 Paper 1, Question 18

Short response · Technology-free · [4 marks]

A person enters the lowest carriage of a miniature Ferris wheel with a six-metre diameter. The bottom carriage is one metre off the ground. When top speed is reached, it takes three seconds for a carriage to travel from the lowest to the highest point of the ride. It is claimed that:

The vertical motion of the Ferris wheel produces a maximum vertical acceleration on each rider that is more than half the acceleration of free fall.

Free fall occurs when gravity is the only force acting, resulting in an acceleration of $9.8\ \text{m\,s}^{-2}$.

Evaluate the reasonableness of the claim.

This one is complex unfamiliar, with its own full walkthrough.

The wheel keeps going round, so the height repeats. That means a periodic model.

1 markRecognising that the motion needs a periodic model. Saying it in words or writing a trig equation both count.

Build the model. The radius is 3 m, so the amplitude is 3. The centre is $1+3=4$ m off the ground. Bottom to top takes 3 s, so a full turn takes 6 s and $b=\frac{2\pi}{6}=\frac{\pi}{3}$. The rider starts at the minimum, so use negative cosine:

$$h(t)=4-3\cos\left(\frac{\pi}{3}t\right)$$

1 markThe model. A sine model with a phase shift is fine too.

Acceleration is the second derivative:

$$h''(t)=\frac{\pi^{2}}{3}\cos\left(\frac{\pi}{3}t\right)$$

The largest it can be is $\frac{\pi^{2}}{3}\approx\frac{9.87}{3}\approx 3.3\ \text{m\,s}^{-2}$.

1 markThe maximum acceleration, from your second derivative. If an earlier answer was wrong, you can still get this mark for using it correctly.

Half of free fall is $4.9\ \text{m\,s}^{-2}$, and $3.3$ is less than that. The claim is not reasonable.

1 markA statement about whether the claim is reasonable.

Answer

Not reasonable. The maximum is about $3.3\ \text{m\,s}^{-2}$, less than $4.9$.

2022 Paper 2, Question 13

Short response · Technology-active · [4 marks]

A sandy beach has a fence on one side and ocean on the other. The width of the beach is the distance (in metres) from the fence to the water's edge. The width, $w(t)$, at a certain point is given by $w(t)=a+b\sin\left(\dfrac{\pi}{6}t-\dfrac{\pi}{3}\right),\ 0\le t\le 24$, where $t$ is time (in hours) since 6 am. The width of the beach is 8 metres at 8 am and 3 metres at 5 pm.

(a) Determine $a$ and $b$.[2 marks]

(b) Determine the rate of change of the width of the beach at 8 am and the first time after this when this rate of change is repeated.[2 marks]

(a) 8 am is $t=2$, where the sine part is $\sin(0)=0$, so $a=8$.

1 markFinding $a$.

5 pm is $t=11$, where it is $\sin\frac{3\pi}{2}=-1$, so $8-b=3$ and $b=5$.

1 markFinding $b$.

(b) Differentiate and substitute $t=2$:

$$w'(t)=\frac{5\pi}{6}\cos\left(\frac{\pi}{6}t-\frac{\pi}{3}\right),\qquad w'(2)=\frac{5\pi}{6}\approx 2.62$$

1 markThe rate at 8 am. If an earlier answer was wrong, you can still get this mark for using it correctly.

The cosine is next equal to 1 one full period later. The period is 12 hours, so that is $t=14$, which is 8 pm.

1 markThe first time the rate repeats. Using a graph, a table or the period all count.

Answer

(a) $a=8,\ b=5$ · (b) $\frac{5\pi}{6}\approx 2.62$ m per hour, repeated at 8 pm

2024 Paper 1, Question 19

Short response · Technology-free · [6 marks]

A permanent ice glacier is in a valley in New Zealand. Due to the temperature changes of the seasons each year, the glacier expands for six months and recedes for six months. The changing distance of a point on the front edge of the glacier to a car park can be modelled by a sine function.

During the colder months, when the glacier expands, the front edge of the glacier moves to within 270 m of the car park. However, in the warmer months, when the glacier recedes, the front edge moves to a maximum distance of 280 m away from the car park.

The erosion effects of the glacier on the ground are of most interest to geologists when the absolute value of the acceleration of the front edge is greater than $\dfrac{5\pi^{2}\sqrt3}{72}$ metres/month². During these times, a team of geologists sets up a camp site nearby to perform field work. Whenever the acceleration is less than this, the geologists leave camp. The following claim is made.

The geologists will spend a total of between seven and eight months at the camp site each calendar year.

Evaluate the reasonableness of this claim.

This one is complex unfamiliar, with its own full walkthrough.

Build the model. The distance runs between 270 and 280 m, so the midline is 275 and the amplitude 5. One expand-and-recede cycle takes 12 months, so $b=\frac{\pi}{6}$. Where it starts does not matter here, because you only need lengths of time:

$$d(t)=275+5\sin\left(\frac{\pi}{6}t\right)$$

1 markThe model.

Differentiate once for the velocity:

$$d'(t)=\frac{5\pi}{6}\cos\left(\frac{\pi}{6}t\right)$$

1 markThe velocity. If an earlier answer was wrong, you can still get this mark for using it correctly.

And again for the acceleration:

$$d''(t)=-\frac{5\pi^{2}}{36}\sin\left(\frac{\pi}{6}t\right)$$

1 markThe acceleration.

Set the size of the acceleration above the threshold and divide through by $\frac{5\pi^{2}}{36}$:

$$\left|\sin\left(\frac{\pi}{6}t\right)\right|>\frac{\sqrt3}{2}$$

On the unit circle, that is true for $\frac{\pi}{6}t$ between $\frac{\pi}{3}$ and $\frac{2\pi}{3}$, so between $t=2$ and $t=4$ months.

1 markOne stretch of time when the acceleration is big enough.

It is also true between $\frac{4\pi}{3}$ and $\frac{5\pi}{3}$, so between $t=8$ and $t=10$. That is $2+2=4$ months a year.

1 markThe total time. Counting each stretch as 3 months, so 6 in total, is also accepted.

The claim was between 7 and 8 months, so it is not reasonable.

1 markDeciding on the claim. This only counts if your working backs it up.

Answer

Not reasonable. They spend 4 months at camp each year.

Game plan

Your game plan for differentiation questions

Ask yourself these, in order, every time. They take you from reading the question to knowing exactly what to do. Try them on any Phase 3 or 4 question above.

1Check that you need to differentiate. The question will want a rate of change.
2What are you asked to find? A maximum, a rate of change, a velocity, a gradient or a tangent line.
3Is your equation ready to differentiate? If not, find the unknown first.
4How many times do you need to differentiate? Your answer to step 2 decides this.
5Which differentiation rules apply here?
6What units is the question in? Give your answer in those units.

Words that mean "differentiate"

rate of change, or instantaneous rate of change
"the rate at which … is changing", "how fast … is changing"
velocity (first derivative) and acceleration (second derivative)
gradient of the tangent, or of the curve, at a point
"increasing or decreasing by [rate] per [unit]"

The clue: if the answer needs a rate unit, like °C per hour or metres per month², you differentiate.

Practice

Practice questions

Eight questions of my own, covering every skill. Each one is labelled with the skill it practises, so if one goes wrong you know which part of the guide to go back to. Try them with the game plan beside you.

Question 1 of 8

Skill 1: the sine rules · Technology-free · [2 marks]

Determine $\dfrac{dy}{dx}$ for $y=3\sin(4x-\pi)$.

1This is $\sin f(x)$ with $f(x)=4x-\pi$, so use $\dfrac{d}{dx}\sin f(x)=f'(x)\cos f(x)$. Here $f'(x)=4$.

2The 3 out the front just multiplies through.

$$\frac{dy}{dx}=3\times 4\cos(4x-\pi)=12\cos(4x-\pi)$$

Leave the bracket exactly as it was. It does not change when you differentiate.

Answer

$\dfrac{dy}{dx}=12\cos(4x-\pi)$

Question 2 of 8

Skill 2: the cosine rules · Multiple choice · Technology-free · [1 mark]

The derivative of $\cos(2x)$ with respect to $x$ is

Use $\dfrac{d}{dx}\cos f(x)=-f'(x)\sin f(x)$ with $f(x)=2x$, so $f'(x)=2$. That gives $-2\sin(2x)$. Option (A) has lost the minus sign, and (C) has forgotten the chain rule.

Answer

(B)

Question 3 of 8

Skill 1: the sine rules, with the product rule · Technology-free · [3 marks]

Determine the derivative of $f(x)=x^{2}\sin(3x)$. Express your answer in factorised form.

1Two functions are multiplied, so you need the product rule. The $\sin(3x)$ also needs the chain rule.

$$u=x^{2}\quad u'=2x\qquad\quad v=\sin(3x)\quad v'=3\cos(3x)$$

2Put the pieces into $f'(x)=u'v+uv'$.

$$f'(x)=2x\sin(3x)+3x^{2}\cos(3x)$$

3Take out the common factor of $x$, because the question asks for factorised form.

Answer

$f'(x)=x\big(2\sin(3x)+3x\cos(3x)\big)$

Question 4 of 8

Skill 2: the cosine rules, with the quotient rule · Technology-free · [3 marks]

Determine the derivative of $y=\dfrac{\cos x}{x}$.

1One function is divided by another, so use the quotient rule.

$$u=\cos x\quad u'=-\sin x\qquad\quad v=x\quad v'=1$$

2Substitute into $\dfrac{u'v-uv'}{v^{2}}$. Keep the minus sign from $u'$ in a bracket so it does not go missing.

$$\frac{dy}{dx}=\frac{(-\sin x)(x)-(\cos x)(1)}{x^{2}}=\frac{-x\sin x-\cos x}{x^{2}}$$

Answer

$\dfrac{dy}{dx}=-\dfrac{x\sin x+\cos x}{x^{2}}$

Question 5 of 8

Skill 2: the cosine rules, with the product rule · Technology-free · [3 marks]

Determine the exact gradient of $f(x)=e^{x}\cos x$ at $x=\dfrac{\pi}{2}$.

1Two functions are multiplied, so use the product rule.

$$u=e^{x}\quad u'=e^{x}\qquad\quad v=\cos x\quad v'=-\sin x$$

$$f'(x)=e^{x}\cos x-e^{x}\sin x=e^{x}(\cos x-\sin x)$$

2Substitute $x=\dfrac{\pi}{2}$. From the unit circle, $\cos\dfrac{\pi}{2}=0$ and $\sin\dfrac{\pi}{2}=1$.

$$f'\!\left(\tfrac{\pi}{2}\right)=e^{\pi/2}(0-1)=-e^{\pi/2}$$

The question says exact, so leave $e^{\pi/2}$ as it is.

Answer

$-e^{\pi/2}$

Question 6 of 8

Skill 2: the cosine rules, with Unit 1 and 2 trig · Technology-free · [3 marks]

Determine the values of $x$, for $0\le x\le 2\pi$, where the gradient of $y=2\cos x+\sqrt{3}\,x$ is zero.

1Differentiate each term.

$$\frac{dy}{dx}=-2\sin x+\sqrt{3}$$

2Set the derivative equal to zero and rearrange.

$$-2\sin x+\sqrt{3}=0\ \Rightarrow\ \sin x=\frac{\sqrt{3}}{2}$$

3From the special triangles, the reference angle is $\dfrac{\pi}{3}$. Sine is positive in the first and second quadrants, so there are two solutions.

$$x=\frac{\pi}{3}\quad\text{or}\quad x=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$$

Answer

$x=\dfrac{\pi}{3}$ and $x=\dfrac{2\pi}{3}$

Question 7 of 8

Skill 3: modelling with a known equation · Technology-free · [3 marks]

The depth of water at a jetty is modelled by $D=3\sin\left(\dfrac{\pi t}{6}\right)+5$, where $D$ is in metres and $t$ is the number of hours after midnight. Determine the rate at which the depth is changing at 2 am.

1Check that you need to differentiate. The question asks for a rate of change.

2What are you asked to find? $\dfrac{dD}{dt}$ at $t=2$.

3Is the equation ready? Yes. There are no unknowns.

4How many times? Once.

5Which rule? $\dfrac{d}{dx}\sin f(x)=f'(x)\cos f(x)$, with $f'(t)=\dfrac{\pi}{6}$.

$$\frac{dD}{dt}=3\times\frac{\pi}{6}\cos\left(\frac{\pi t}{6}\right)=\frac{\pi}{2}\cos\left(\frac{\pi t}{6}\right)$$

$$\text{At } t=2:\quad \frac{dD}{dt}=\frac{\pi}{2}\cos\left(\frac{\pi}{3}\right)=\frac{\pi}{2}\times\frac{1}{2}=\frac{\pi}{4}$$

6Units. The depth is in metres and time is in hours, so the rate is in metres per hour.

Answer

The depth is increasing at $\dfrac{\pi}{4}$ m/h (about 0.79 m/h)

Question 8 of 8

Skill 3: modelling with an unknown equation · Technology-active · [4 marks]

On a spring day the temperature in a town reaches a minimum of 18°C at 3 am and a maximum of 30°C at 3 pm. Assume the temperature follows a trigonometric model with a period of 24 hours, and let $t$ be the number of hours after 3 am.

(a) Determine a model for the temperature $T$ (°C). [2 marks]

(b) Determine the rate at which the temperature is changing at 9 am. [2 marks]

(a) The model starts at a minimum, so a negative cosine fits and you do not need a horizontal shift.

$$a=\frac{30-18}{2}=6\qquad k=\frac{30+18}{2}=24\qquad b=\frac{2\pi}{24}=\frac{\pi}{12}$$

$$T=-6\cos\left(\frac{\pi t}{12}\right)+24$$

Check it: at $t=0$, $T=-6+24=18$. At $t=12$, $T=-6\cos\pi+24=30$.

(b) Now the equation is ready, so work through the game plan.

1Check that you need to differentiate. The question asks for a rate of change.

2What are you asked to find? $\dfrac{dT}{dt}$ at 9 am, which is $t=6$. Be careful here. $t$ counts from 3 am. It does not count from midnight.

3Is the equation ready? Yes, from part (a).

4How many times? Once.

5Which rule? $\dfrac{d}{dx}\cos f(x)=-f'(x)\sin f(x)$. The two minus signs make a plus.

$$\frac{dT}{dt}=6\times\frac{\pi}{12}\sin\left(\frac{\pi t}{12}\right)=\frac{\pi}{2}\sin\left(\frac{\pi t}{12}\right)$$

$$\text{At } t=6:\quad \frac{dT}{dt}=\frac{\pi}{2}\sin\left(\frac{\pi}{2}\right)=\frac{\pi}{2}\approx 1.57$$

6Units. Degrees per hour.

Answer

(a) $T=-6\cos\left(\frac{\pi t}{12}\right)+24$ (b) increasing at about 1.57°C per hour

Keep practising

Ready for something harder?

Both complex unfamiliar questions from Phase 4 have full walkthroughs. Start with the glacier: you build a sine model from a description, then use the unit circle to turn an acceleration threshold into a length of time.

Open 2024 Paper 1, Question 19 Open 2023 Paper 1, Question 18

Past questions and marking-guide steps are from the 2020–2025 QCAA Mathematical Methods external assessments, © State of Queensland (QCAA) 2020–2025, licensed under CC BY 4.0, and have been adapted. Practice questions are written by Tangent Tuition. Tangent Tuition is not affiliated with the QCAA.

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