Mathematical Methods · Units 3 & 4

Derivatives of Circular Functions

Q: What is the derivative of sin, cos and tan?

The derivative of sine is cosine, the derivative of cosine is negative sine, and the derivative of tangent is secant squared, which equals one over cosine squared. These three results only hold when the angle is measured in radians.

Q: Why do circular function derivatives only work in radians, not degrees?

The neat rules depend on the angle being in radians. If your calculator is set to degrees the slope of sine is no longer simply cosine, because changing the units of the input changes how fast the output rises and falls. Every VCAA exam uses radians, so always work in radians and exact values like pi over three.

Q: How do I differentiate sin(kx) or cos of something with a number in front?

Use the chain rule. Differentiate the outside, copying the bracket down unchanged, then multiply by the derivative of the inside. For sin(kx) that gives k cos(kx). Any constant sitting out the front just rides along and multiplies the answer, and a phase shift inside the bracket adds nothing to the inside derivative.

Learn how to differentiate sine, cosine and tangent the easy way, with plain English intuition, the chain rule for sin(kx), worked examples and an auto marked practice test. VCE Maths Methods Units 3 and 4.

Waves are everywhere. Sound, light, tides, the swing of a pendulum and the bob of a spring all rise and fall in the smooth shape of a sine curve. Differentiating a circular function is how you find the rate at which a wave is rising or falling at any instant. The wonderful surprise is that the slope of a wave is just another wave, shifted along a little. Learn three short results and a single rule, and you can find the steepness of any sine, cosine or tangent in seconds.

The whole topic in three chunks

1 · The three base derivatives

$\dfrac{d}{dx}\sin x = \cos x$
$\dfrac{d}{dx}\cos x = -\sin x$ (cosine picks up a minus)
$\dfrac{d}{dx}\tan x = \sec^2 x = \dfrac{1}{\cos^2 x}$

2 · Radians only

These neat rules hold only when the angle is in radians
In degrees the slope of $\sin x$ is no longer $\cos x$
Think in $\pi$ and exact values like $\dfrac{\pi}{3}$ , never $60^\circ$

3 · The chain rule for sin(kx)

Differentiate the outside, copy the bracket down unchanged
Multiply by the inside derivative: $\dfrac{d}{dx}\sin(kx) = k\cos(kx)$
A constant out the front just rides along

The three results to memorise

A circular function is just a fancy name for $\sin$ , $\cos$ and $\tan$ . The whole topic rests on three facts about their gradients. In radians (and circular function calculus only works in radians, never degrees) the rules are beautifully neat.

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

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

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

Notice the pattern. The slope of $\sin$ is $\cos$ . The slope of $\cos$ is $\sin$ but flipped negative. That minus sign is the most forgotten thing in the whole topic, so say it out loud every time. Cosine picks up a minus.

The blue curve y = sin(x) and its derivative y = cos(x) (red, dashed): where sin is steepest (at x = 0) cos peaks at 1, and where sin levels off at its crests cos crosses zero, showing the slope of a wave is just another wave shifted along.

Why these only work in radians

The neat rules above are a gift, but the gift only arrives if your angle is measured in radians. If you ever set your calculator to degrees, the slope of $\sin x$ is no longer simply $\cos x$ , because the units of the input change how fast the output moves. This is exactly why every VCAA exam, and every question on this page, uses radians and exact values like $\dfrac{\pi}{3}$ rather than $60$ degrees. Train yourself to think in $\pi$ from the start.

Stretching the wave: sin(kx) and the chain rule

Real waves rarely come as a plain $\sin x$ . They get squashed or stretched, written as $\sin(kx)$ where $k$ controls how many cycles fit in. To differentiate these you bolt the chain rule onto the three results above. The inside function is whatever sits in the brackets, and its derivative gets multiplied out the front.

If $y = \sin(kx)$ , treat $kx$ as the inside. Differentiate the outside to get $\cos(kx)$ , then multiply by the derivative of the inside, which is $k$ :

\frac{d}{dx}\big(\sin(kx)\big) = k\cos(kx)

The same idea gives the matching cosine and tangent rules. The factor $k$ always lands out the front, and the bracket is copied down completely unchanged.

\frac{d}{dx}\big(\cos(kx)\big) = -k\sin(kx), \qquad \frac{d}{dx}\big(\tan(kx)\big) = k\sec^2(kx)

This works for any inside, not just $kx$ . For $y = \sin(2x + \tfrac{\pi}{6})$ the inside is $2x + \tfrac{\pi}{6}$ , whose derivative is still just $2$ , so the answer is $2\cos(2x + \tfrac{\pi}{6})$ . The phase shift $\tfrac{\pi}{6}$ is a constant, so it adds nothing to the inside derivative and stays sitting inside the bracket.

The method, step by step

Every circular derivative in the course follows the same short recipe. The skill is doing it cleanly without dropping a factor or a sign.

Identify the inside function, which is everything inside the bracket.
Differentiate the outside, using $\sin \to \cos$ , $\cos \to -\sin$ , or $\tan \to \sec^2$ , copying the inside down unchanged.
Multiply by the derivative of the inside.
Multiply by any constant sitting out the front, and tidy the signs.

The examiner reports flag the same handful of errors year after year. Many students omit the inside derivative, turning $\sin(4x)$ into $\cos(4x)$ instead of $4\cos(4x)$ . Many lose the minus sign when differentiating cosine. And some alter the argument, wrongly shrinking $\cos(3x)$ back to $\cos(x)$ . The inside of the bracket never changes when you differentiate. Bring the factor out, keep the bracket the same, fix the sign.

Learn it, then fade the help

1 · I do — fully worked

Differentiate $y = \sin(3x)$ . The inside is $3x$ , whose derivative is $3$ . Differentiate the outside ( $\sin \to \cos$ ), copy the bracket down, then multiply by the inside derivative: $\dfrac{dy}{dx} = \cos(3x) \times 3 = 3\cos(3x)$ .

2 · We do — your turn, guided

Same steps, your turn. Differentiate $y = \sin(5x)$ . The inside is $5x$ , so begin with $\dfrac{dy}{dx} = \cos(5x) \times \;?\;$ .

Check your step

The inside derivative is $5$ , so $\dfrac{dy}{dx} = 5\cos(5x)$ .

3 · You do — solo

No scaffolding now: differentiate $y = \cos(2x)$ . Mind the sign.

Reveal solution

The inside $2x$ has derivative $2$ , and $\cos$ differentiates to $-\sin$ , so $\dfrac{dy}{dx} = -\sin(2x) \times 2 = -2\sin(2x)$ . The minus comes from differentiating cosine.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

What is

\dfrac{d}{dx}\sin x

, and

\dfrac{d}{dx}\cos x

What is

\dfrac{d}{dx}\tan x

Why must the angle be in radians?

Using the chain rule, what is

\dfrac{d}{dx}\sin(kx)

For

f(x) = \tan(5x)

, what is

f'(x)

Recall · Chain Rule

State the chain rule for

\dfrac{dy}{dx}

when

y = f(u)

and

u = g(x)

Recall · Antidifferentiation

What is

\displaystyle\int \cos x \,dx

See this recipe in action in the Worked Examples tab, then test yourself in Try It.

Worked Example 1An online sales model, from a real exam

Online monthly sales are modelled by $f : (0, 36] \to R$ , $f(t) = 3000 + 30t + 700\cos\left(\dfrac{\pi t}{6}\right) + 400\cos\left(\dfrac{\pi t}{3}\right)$ . (i) The model predicts that every 12 months monthly sales increase by $n$ million dollars; find $n$ . (ii) Find $f'(t)$ . (iii) Hence find the maximum instantaneous rate of change of $f$ , to the nearest million dollars per month, and the values of $t$ in $(0, 36]$ at which it occurs, to one decimal place.

1
Both cosine terms have period 12, so over 12 months only the linear term changes. The increase is $30 \times 12$ .
$n = 30 \times 12 = 360$
2
Differentiate term by term. Each cosine brings down its inside derivative, $\frac{\pi}{6}$ or $\frac{\pi}{3}$ , and turns into a negative sine.
$f'(t) = 30 - \frac{350\pi}{3}\sin\left(\frac{\pi t}{6}\right) - \frac{400\pi}{3}\sin\left(\frac{\pi t}{3}\right)$
3
Maximise $f'$ with technology. Since $f'$ has period 12, the maximum recurs three times across $(0, 36]$ .
$\text{max} \approx 725,\quad t \approx 10.2,\ 22.2,\ 34.2$

Answer

n = 360;\ f'(t) = 30 - \dfrac{350\pi}{3}\sin\left(\dfrac{\pi t}{6}\right) - \dfrac{400\pi}{3}\sin\left(\dfrac{\pi t}{3}\right);\ \text{max rate} \approx 725 \text{ at } t \approx 10.2, 22.2, 34.2

VCAA 2024 Mathematical Methods Exam 2, Section B Q3b

Worked Example 2The range of a derivative on an interval, from a real exam

For $f'(x) = \sin(x) + x\cos(x)$ , determine the range of $f'(x)$ over the interval $\left[\dfrac{\pi}{2}, \dfrac{2\pi}{3}\right]$ .

1
Evaluate $f'$ at the left endpoint. Here $\cos\frac{\pi}{2} = 0$ , so the second term vanishes.
$f'\left(\tfrac{\pi}{2}\right) = \sin\tfrac{\pi}{2} + \tfrac{\pi}{2}\cos\tfrac{\pi}{2} = 1$
2
Evaluate $f'$ at the right endpoint, using $\sin\frac{2\pi}{3} = \frac{\sqrt3}{2}$ and $\cos\frac{2\pi}{3} = -\frac12$ .
$f'\left(\tfrac{2\pi}{3}\right) = \frac{\sqrt3}{2} + \frac{2\pi}{3}\left(-\tfrac12\right) = \frac{\sqrt3}{2} - \frac{\pi}{3}$
3
$f'$ is decreasing across this interval, so the range runs from the smaller right-endpoint value up to $1$ . Use square brackets, smaller value first.
$\text{Range} = \left[\frac{\sqrt3}{2} - \frac{\pi}{3},\ 1\right]$

Answer

\text{Range} = \left[\dfrac{\sqrt3}{2} - \dfrac{\pi}{3},\ 1\right]

VCAA 2024 Mathematical Methods Exam 1, Q7b.ii

Worked Example 3A scaled sine wave

Differentiate $y = \sin(3x)$ .

1
Spot the inside function. Here the inside is $3x$ , so use the chain rule.
$y = \sin(u), \quad u = 3x$
2
Differentiate the outside. The derivative of $\sin$ is $\cos$ , with the inside left alone.
$\frac{dy}{du} = \cos(u) = \cos(3x)$
3
Differentiate the inside and multiply. The derivative of $3x$ is $3$ .
$\frac{du}{dx} = 3$

Answer

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

Worked Example 4Cosine with a coefficient and a phase

Differentiate $y = 4\cos\left(2x - \dfrac{\pi}{3}\right)$ .

1
The front number $4$ just rides along. The inside is $2x - \frac{\pi}{3}$ , whose derivative is $2$ .
$\frac{d}{dx}\cos(2x - \tfrac{\pi}{3}) = -\sin\left(2x - \tfrac{\pi}{3}\right) \times 2$
2
Multiply by the front $4$ and the inside derivative $2$ . Watch the sign: $\cos$ differentiates to $-\sin$ .
$\frac{dy}{dx} = 4 \times (-\sin(2x - \tfrac{\pi}{3})) \times 2$

Answer

\frac{dy}{dx} = -8\sin\left(2x - \dfrac{\pi}{3}\right)

Worked Example 5A tangent, then evaluate

For $f(x) = \tan(5x)$ , find $f'\left(\dfrac{\pi}{20}\right)$ .

1
The derivative of $\tan(u)$ is $\sec^2(u)$ , so the chain rule gives the inside derivative $5$ out the front.
$f'(x) = 5\sec^2(5x) = \frac{5}{\cos^2(5x)}$
2
Substitute $x = \frac{\pi}{20}$ , so the inside becomes $5 \times \frac{\pi}{20} = \frac{\pi}{4}$ .
$f'\left(\tfrac{\pi}{20}\right) = \frac{5}{\cos^2(\frac{\pi}{4})}$
3
Use the exact value $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ , so $\cos^2\frac{\pi}{4} = \frac{1}{2}$ .
$f'\left(\tfrac{\pi}{20}\right) = \frac{5}{\frac{1}{2}} = 10$

Answer

f'\left(\dfrac{\pi}{20}\right) = 10

Practice test

Try it yourself

7 questions, 9 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.The function $f$ is given by $f(x) = \tan\left(\dfrac{x}{2}\right)$ for $4 \le x < 2\pi$ and $f(x) = \sin(ax)$ for $2\pi \le x \le 8$ . The value of $a$ for which $f$ is continuous and smooth at $x = 2\pi$ is:

1mark

Need a hint?

Continuity forces

\sin(2\pi a) = 0

. Smoothness forces the two derivatives to match at

x = 2\pi

; the gradient of

\tan(x/2)

there is

\frac12

VCAA 2023 Mathematical Methods Exam 2, Section A Q9

Q2.The derivative of $y = \sin(4x)$ is:

1mark

Need a hint?

Differentiate the outside (

\sin \to \cos

), then multiply by the derivative of the inside

4x

Q3.If $f(x) = \cos(2x)$ , then $f'(x)$ equals:

1mark

Need a hint?

Cosine picks up a minus sign when you differentiate it, and the inside

2x

still contributes a factor.

Q4.The gradient of $y = 3\sin\left(2x + \dfrac{\pi}{6}\right)$ at any point is given by $\dfrac{dy}{dx} =$

1mark

Need a hint?

The front

3

rides along and the phase

\frac{\pi}{6}

is a constant, so only

2x

matters for the inside derivative.

Q5.For $f(x) = \tan(3x)$ , the value of $f'\left(\dfrac{\pi}{12}\right)$ is:

1mark

Need a hint?

First differentiate

\tan(3x)

3\sec^2(3x)

, then substitute. Note

3 \times \frac{\pi}{12} = \frac{\pi}{4}

Q6.The derivative of $y = \cos\left(\dfrac{x}{2}\right)$ is:

1mark

Need a hint?

Rewrite the inside as

\frac{1}{2}x

and ask what its derivative is before multiplying.

Q7.Let $f(x) = 2\sin(3x)$ . Find the gradient of the graph of $f$ at the point where $x = \dfrac{\pi}{9}$ . Give the exact value and show all working.

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

What is the derivative of sin, cos and tan?

The derivative of sine is cosine, the derivative of cosine is negative sine, and the derivative of tangent is secant squared, which equals one over cosine squared. These three results only hold when the angle is measured in radians.

Why do circular function derivatives only work in radians, not degrees?

The neat rules depend on the angle being in radians. If your calculator is set to degrees the slope of sine is no longer simply cosine, because changing the units of the input changes how fast the output rises and falls. Every VCAA exam uses radians, so always work in radians and exact values like pi over three.

How do I differentiate sin(kx) or cos of something with a number in front?

Use the chain rule. Differentiate the outside, copying the bracket down unchanged, then multiply by the derivative of the inside. For sin(kx) that gives k cos(kx). Any constant sitting out the front just rides along and multiplies the answer, and a phase shift inside the bracket adds nothing to the inside derivative.

Derivatives of Circular Functions

Learn

The three results to memorise

Why these only work in radians

Stretching the wave: sin(kx) and the chain rule

The method, step by step

Lock it in with active recall

Worked examples

Practice questions

Try it yourself

Frequently asked questions