MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: derivatives of trigonometric functions (MATH 140)

Q: Why must x be in radians to differentiate sin x?

Because the formula comes from the limit of sin h over h as h goes to zero, and that limit equals 1 only in radians: its proof uses the area of a circular sector, one half of r squared times the angle, which is a radian formula. In degrees the same limit is pi over 180, and the derivative of the sine of x degrees is pi over 180 times its cosine.

Q: How do I remember the signs of the six trig derivatives?

The three co-functions, cosine, cotangent and cosecant, have a minus sign in their derivatives, and the three others do not. Each co-derivative is also the mirror of its partner: swap every function with its co-function and add the minus. Tangent gives secant squared, so cotangent gives minus cosecant squared; secant gives secant tangent, so cosecant gives minus cosecant cotangent.

Q: How do I compute the limit of sin 5x over 3x as x goes to 0?

Make the same expression appear under the sine and in the denominator. Write the quotient as five thirds times sin 5x over 5x. Since 5x goes to zero, sin 5x over 5x goes to 1, and the limit is five thirds. Answering 1 directly is the most common mistake: the rule needs the same box in both places.

Q: Can I differentiate sin 2x before learning the chain rule?

Yes. Write sin 2x as 2 sin x cos x and use the product rule: the derivative is 2 times cos squared x minus sin squared x, which is 2 cos 2x by the double angle formula. The factor 2 is the one students drop when they answer cos 2x. The chain rule of the next chapter gives the same result mechanically.

Q: What is the 99th derivative of sin x?

The derivatives of sine repeat every four steps: cosine, minus sine, minus cosine, then sine again. Divide 99 by 4: the remainder is 3, so the 99th derivative equals the third one, minus cosine of x. Reasoning only on whether 99 is odd tells you it is a cosine, but not its sign.

This sheet is not a summary of section 3.3 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on the derivatives of trigonometric functions in MATH 140 at McGill University, and which precise gesture avoids each loss.

The chapter looks like a table of six formulas to memorize, and the formulas are rarely the problem. The marks are lost on the limit behind them, on a minus sign, on a quotient differentiated as if the rule did not exist, and on an equation $f'(x) = 0$ solved by dividing. Every value below is exact and done by hand, as on the exam.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next and tells you which chapter to tackle next. Create your space, an email is enough.

The thread of the chapter

Every trigonometric derivative is borrowed from ONE limit, $\frac{\sin\theta}{\theta} \to 1$ , and inherits its three conditions: the angle in RADIANS, the variable going to $0$ , the SAME quantity under the sine and in the denominator; beyond it, the product and quotient rules applied honestly and the minus sign of every co-function.

This chapter is part of MATH 140, Calculus 1 (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Fundamental trigonometric limit
Trigonometric limits
Radian measure
Derivative of sine and cosine
Derivatives of trigonometric functions
Product rule
Quotient rule
Pythagorean identity
Higher-order derivatives
Tangent line
Horizontal tangent line
Simple harmonic motion

The essentials

One limit, three conditions

• $\lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1$ , proved by squeezing $\cos\theta < \frac{\sin\theta}{\theta} < 1$ , which comes from comparing the areas $\frac{1}{2}\sin\theta < \frac{\theta}{2} < \frac{1}{2}\tan\theta$ on the unit circle.
• Condition 1, RADIANS: the sector area $\frac{\theta}{2}$ is a radian formula. In degrees, $\frac{\sin(x^\circ)}{x} \to \frac{\pi}{180}$ .
• Condition 2, the variable goes to $0$ : at infinity $\frac{\sin x}{x} \to 0$ by the squeeze, and at $\pi$ the quotient $\frac{\sin x}{x}$ is simply $0$ .
• Condition 3, the SAME box: $\frac{\sin\square}{\square} \to 1$ only if the expression under the sine is the one in the denominator. $\frac{\sin 3x}{x} = 3\cdot\frac{\sin 3x}{3x} \to 3$ .
• The companion: $\frac{1 - \cos\theta}{\theta} = \frac{\sin\theta}{\theta}\cdot\frac{\sin\theta}{1 + \cos\theta} \to 0$ , and $\frac{1 - \cos\theta}{\theta^2} \to \frac{1}{2}$ .

Both quotients have no value at

0

, but the blue one tends to

1

and the orange one to

0

1 - \cos x

shrinks much faster than

x

The proof of $(\sin x)' = \cos x$ uses exactly these two limits, with $h$ as the variable: a limit written with the wrong value, $1$ for $\frac{1 - \cos h}{h}$ , gives a wrong derivative.

The six derivatives, and where each sign comes from

• $(\sin x)' = \cos x$ and $(\cos x)' = -\sin x$ , both from the definition and the addition formulas; the minus comes from $\cos(x + h) = \cos x\cos h - \sin x\sin h$ .
• $(\tan x)' = \sec^2 x = 1 + \tan^2 x$ and $(\sec x)' = \sec x\tan x$ , by the quotient rule on $\frac{\sin x}{\cos x}$ and $\frac{1}{\cos x}$ , where $\cos x \ne 0$ .
• $(\cot x)' = -\csc^2 x$ and $(\csc x)' = -\csc x\cot x$ , where $\sin x \ne 0$ . The three CO-functions take a minus sign.
• The derivatives of $\sin$ and $\cos$ repeat every four steps: $\sin \to \cos \to -\sin \to -\cos \to \sin$ .
• Read $\cos x$ as the slope of $\sin x$ : slope $1$ at $0$ , $0$ at $\frac{\pi}{2}$ , $-1$ at $\pi$ . It is the fastest check of a sign.

Where

\sin x

is flat, at

\frac{\pi}{2}

and

\frac{3\pi}{2}

\cos x

crosses zero; where

\sin x

climbs steepest, at

0

and

2\pi

\cos x

is at its top value

1

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The six derivatives, each with a slope you can check

Read a line as: the function of the first column, rewritten as in the second, has the derivative of the third. The red lines are rules that do not exist: two students in three write one of them at least once in the term.

Function	Rewritten as	Derivative
$\sin x$	definition	$\cos x$
Example: Slope at $0$ : $\cos 0 = 1$ , so the tangent at the origin is $y = x$ .
$\cos x$	definition	$-\sin x$
Example: Slope at $\frac{\pi}{2}$ : $-\sin\frac{\pi}{2} = -1$ ; cosine is falling there, through $0$ .
$\tan x$	$\frac{\sin x}{\cos x}$	$\sec^2 x$
Example: Slope at $\frac{\pi}{4}$ : $\sec^2\frac{\pi}{4} = (\sqrt 2)^2 = 2$ .
$\sec x$	$\frac{1}{\cos x}$	$\sec x\tan x$
Example: Slope at $\frac{\pi}{3}$ : $2\cdot\sqrt 3 = 2\sqrt 3$ .
$\cot x$	$\frac{\cos x}{\sin x}$	$-\csc^2 x$
Example: Slope at $\frac{\pi}{4}$ : $-\csc^2\frac{\pi}{4} = -2$ , the mirror of $\tan x$ .
$\csc x$	$\frac{1}{\sin x}$	$-\csc x\cot x$
Example: Slope at $\frac{\pi}{6}$ : $-2\cdot\sqrt 3 = -2\sqrt 3$ .
$\tan x$	$\frac{(\sin x)'}{(\cos x)'}$	$-\cot x$ rule that does not exist
Example: At $\frac{\pi}{4}$ it gives $-1$ , while the true slope is $2$ ; at $0$ it is not even defined. What to do: Quotient rule: $\frac{\cos x\cos x - \sin x(-\sin x)}{\cos^2 x} = \frac{1}{\cos^2 x}$ .
$\sin x\cos x$	$(\sin x)'(\cos x)'$	$-\sin x\cos x$ rule that does not exist
Example: At $0$ it gives $0$ , while the true slope is $\cos^2 0 - \sin^2 0 = 1$ . What to do: Product rule: $\cos x\cos x + \sin x(-\sin x) = \cos^2 x - \sin^2 x$ .

Swap each function with its co-function and add a minus: $\tan \to \sec^2$ becomes $\cot \to -\csc^2$ , and $\sec \to \sec\tan$ becomes $\csc \to -\csc\cot$ .

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Using the standard limit with the wrong box

2 marks, the whole limit

What not to write

“ $\lim_{x\to 0}\frac{\sin 3x}{x} = 1$ , since $\frac{\sin\square}{\square} \to 1$ .”

What to write

“ $\frac{\sin 3x}{x} = 3\cdot\frac{\sin 3x}{3x}$ , and $3x \to 0$ , so the limit is $3 \cdot 1 = 3$ .”

Why: The rule needs the SAME expression under the sine and in the denominator. Multiply and divide by the constant that creates it, before any limit law.

2. Answering 1 to one minus cosine over x

2 marks, and a wrong derivative of sine if it is used in the proof

What not to write

“ $\lim_{x\to 0}\frac{1 - \cos x}{x} = 1$ , like $\frac{\sin x}{x}$ .”

What to write

“ $\frac{1 - \cos x}{x} = \frac{\sin x}{x}\cdot\frac{\sin x}{1 + \cos x} \to 1 \cdot \frac{0}{2} = 0$ .”

Why: $1 - \cos x$ behaves like $\frac{x^2}{2}$ , not like $x$ : it is $\frac{1 - \cos x}{x^2}$ that tends to a nonzero number, $\frac{1}{2}$ . The figure of the first block shows the two curves.

3. Replacing sin x and tan x by x inside a difference

the whole question

What not to write

“ $\tan x \approx x$ and $\sin x \approx x$ , so $\frac{\tan x - \sin x}{x^3} \to \frac{x - x}{x^3} = 0$ .”

What to write

“ $\tan x - \sin x = \frac{\sin x(1 - \cos x)}{\cos x}$ , so the quotient is $\frac{\sin x}{x}\cdot\frac{1 - \cos x}{x^2}\cdot\frac{1}{\cos x} \to 1 \cdot \frac{1}{2} \cdot 1 = \frac{1}{2}$ .”

Why: A limit law applies to a product or a quotient of factors that each have a limit, never to a piece of a difference. Factor until only the standard quotients appear.

4. Losing the minus sign of a co-function

1 mark per derivative, and every tangent line built on it

What not to write

“ $(\csc x)' = \csc x\cot x$ , just as $(\sec x)' = \sec x\tan x$ .”

What to write

“ $(\csc x)' = \left(\frac{1}{\sin x}\right)' = \frac{-\cos x}{\sin^2 x} = -\csc x\cot x$ .”

Why: The co-functions, $\cos$ , $\cot$ and $\csc$ , carry a minus in their derivatives. Sign test: $\csc x$ decreases on $\left(0, \frac{\pi}{2}\right)$ while $\csc x\cot x > 0$ there.

5. Differentiating a quotient as the quotient of the derivatives

the whole derivative

What not to write

“ $(\tan x)' = \frac{(\sin x)'}{(\cos x)'} = \frac{\cos x}{-\sin x} = -\cot x$ .”

What to write

“By the quotient rule, $(\tan x)' = \frac{\cos x\cos x - \sin x(-\sin x)}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$ .”

Why: There is no quotient-of-derivatives rule. The student's answer is not even defined at $0$ , where $y = \tan x$ has the tangent $y = x$ .

6. Differentiating a product as the product of the derivatives

the whole derivative

What not to write

“ $(x^2\sin x)' = 2x\cos x$ .”

What to write

“With $u = x^2$ and $v = \sin x$ : $(uv)' = u'v + uv' = 2x\sin x + x^2\cos x$ .”

Why: Writing $u$ , $v$ , $u'$ and $v'$ on the copy before assembling is the gesture that makes the error impossible, and it is where the method mark is.

7. Dividing by sin x when solving f'(x) = 0

3 of the 5 points, often half the question

What not to write

“ $g(x) = \cos x + \sin^2 x$ : $-\sin x + 2\sin x\cos x = 0$ , so $2\cos x = 1$ , so $x = \frac{\pi}{3}$ or $\frac{5\pi}{3}$ .”

What to write

“ $g'(x) = \sin x\,(2\cos x - 1) = 0$ , so $\sin x = 0$ or $\cos x = \frac{1}{2}$ : $x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi$ .”

Five horizontal tangents on

[0, 2\pi]

: the green ones come from

\cos x = \frac{1}{2}

, the three red ones from

\sin x = 0

, and dividing by

\sin x

erases the red ones.

Why: Dividing by $\sin x$ is legitimate only where $\sin x \ne 0$ , and the lost solutions are exactly where it vanishes. Factor, then set each factor to zero.

8. Differentiating with the angle in degrees

1 mark per derivative, and every numerical slope

What not to write

“If $x$ is in degrees, the derivative of $\sin x$ is still $\cos x$ .”

What to write

“ $\frac{d}{dx}\sin(x^\circ) = \frac{\pi}{180}\cos(x^\circ)$ ; the formula $(\sin x)' = \cos x$ is a statement about radians.”

Why: The proof uses $\frac{\sin h}{h} \to 1$ , which comes from the sector area $\frac{\theta}{2}$ , a radian formula. In degrees the limit is $\frac{\pi}{180}$ and it multiplies every derivative.

9. Reading a high derivative on parity

the whole question, it is only the sign

What not to write

“ $99$ is odd, so $\frac{d^{99}}{dx^{99}}\sin x = \cos x$ .”

What to write

“ $99 = 4 \times 24 + 3$ , so $\frac{d^{99}}{dx^{99}}\sin x = (\sin x)''' = -\cos x$ .”

Why: Parity tells sine from cosine; the SIGN needs the remainder modulo $4$ , because the cycle is $\sin, \cos, -\sin, -\cos$ .

Which method to choose

Which tool for a trigonometric limit, by the FORM of the expression

Look at where the variable goes and at the shape of the expression before writing anything

If $x \to 0$ and $\frac{\sin(ax)}{bx}$ or $\frac{\sin(ax)}{\sin(bx)}$ → make the box appear: multiply and divide by the constant, then use the standard limit

Example: $\frac{\sin 5x}{3x} = \frac{5}{3}\cdot\frac{\sin 5x}{5x} \to \frac{5}{3}$
If a $\tan$ , $\cot$ , $\sec$ or $\csc$ → rewrite in sines and cosines; a cosine going to $1$ is harmless

Example: $\frac{\tan 4x}{x} = 4\cdot\frac{\sin 4x}{4x}\cdot\frac{1}{\cos 4x} \to 4$
If a factor $1 - \cos(\cdot)$ → multiply by the conjugate $1 + \cos(\cdot)$ and use $1 - \cos^2 = \sin^2$

Example: $\frac{1 - \cos x}{x\sin x} = \frac{\sin x}{x}\cdot\frac{1}{1 + \cos x} \to \frac{1}{2}$
If a difference of trigonometric terms, such as $\tan x - \sin x$ → factor into a product of standard quotients; never replace a term by x

Example: $\frac{\tan x - \sin x}{x^3} \to \frac{1}{2}$
If $x \to a \ne 0$ with the form $\frac{0}{0}$ → shift the box to $u = x - a$ , or recognize the difference quotient of a derivative

Example: $\frac{\cos x + 1}{x - \pi} = \frac{\cos x - \cos\pi}{x - \pi} \to -\sin\pi = 0$
If $x \to \pm\infty$ → squeeze with $|\sin| \le 1$ and $|\cos| \le 1$ ; the standard limit says nothing

Example: $-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$ , so the limit is $0$

L'Hospital's rule is not available at this point of MATH 140, and it would be circular here anyway: it needs $(\sin x)' = \cos x$ , which is proved FROM $\frac{\sin x}{x} \to 1$ .

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Proving the derivative of cosine from the definition

When to use it: Any question that says “from the definition” or “using the limit definition of the derivative” about $\sin x$ or $\cos x$

1 Write the difference quotient $\frac{\cos(x + h) - \cos x}{h}$ , and say that $x$ is fixed while $h \to 0$ .
2 Open $\cos(x + h)$ with the addition formula, naming it: $\cos x\cos h - \sin x\sin h$ .
3 Regroup by the constants $\cos x$ and $\sin x$ : $\cos x\cdot\frac{\cos h - 1}{h} - \sin x\cdot\frac{\sin h}{h}$ .
4 Quote the two limits, with $h$ in radians: $\frac{\cos h - 1}{h} \to 0$ and $\frac{\sin h}{h} \to 1$ .
5 Conclude by the sum and product laws: $\cos x \cdot 0 - \sin x \cdot 1 = -\sin x$ .

Concluding sentence

“Since $\lim_{h\to 0}\frac{\cos h - 1}{h} = 0$ and $\lim_{h\to 0}\frac{\sin h}{h} = 1$ , the limit laws give $(\cos x)' = \cos x\cdot 0 - \sin x\cdot 1 = -\sin x$ for every real $x$ .”

The trap: Writing $\lim_{h\to 0}\frac{\cos(x + h) - \cos x}{h} = -\sin x$ “by the table”: a proof from the definition cannot quote the formula it is proving.

Marking: Typically 1 mark for the quotient, 2 for the addition formula, 2 for the regrouping, 3 for the two named limits, 2 for the conclusion.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The slope at 0

Evaluate your derivative at x = 0 and compare with what the graph says there. Sine and tangent leave the origin along y = x, slope 1; cosine is flat at its top, slope 0.

$(\tan x)' = -\cot x$ fails at once: it is undefined at $0$ , while the slope of $\tan x$ there is $1$ .
Parity

The derivative of an even function is odd, and the derivative of an odd function is even. Test your answer at x and at minus x.

$\tan x$ is odd, so its derivative must be even: $\sec^2 x$ is, $-\cot x$ is not.
The sign on an interval

Where you know from the graph that the function falls, the derivative must be negative. One point is enough to catch a lost minus sign.

$\csc x$ falls on $\left(0, \frac{\pi}{2}\right)$ ; at $\frac{\pi}{4}$ , $-\csc x\cot x = -\sqrt 2 < 0$ , while $+\csc x\cot x$ would be positive.
Two forms, one special angle

After simplifying with an identity, evaluate the raw form and the simplified form at a convenient angle such as 0 or pi over 3. They must agree.

$\frac{\cos x + \cos^2 x + \sin^2 x}{(1 + \cos x)^2}$ and $\frac{1}{1 + \cos x}$ both give $\frac{1}{2}$ at $x = 0$ .

The typical problem, taken apart

A quotient that collapses: limit, derivative and tangent line

Let $f(x) = \frac{1 - \cos x}{\sin x}$ on $(0, \pi)$ . Find $\lim_{x\to 0^+} f(x)$ , compute $f'(x)$ in its simplest form, and give the tangent line at $x = \frac{\pi}{2}$ .

No calculator. Name every limit and every rule, as on a MATH 140 final.

The graph starts from a hole at the origin and climbs toward the dashed line

x = \pi

, where

\sin x

vanishes and

1 - \cos x

does not.

Step 1

The form is $\frac{0}{0}$ . Write $f(x) = \frac{1 - \cos x}{x}\cdot\frac{x}{\sin x}$ : the first factor tends to $0$ , the second to $1$ , so $\lim_{x\to 0^+} f(x) = 0$ .

Why

The two standard limits appear with the same box $x$ . Splitting as $(1 - \cos x)\cdot\frac{1}{\sin x}$ instead gives the form $0 \cdot \infty$ and nothing.

Step 2

Quotient rule with $u = 1 - \cos x$ , $u' = \sin x$ , $v = \sin x$ , $v' = \cos x$ : $f'(x) = \frac{\sin x\cdot\sin x - (1 - \cos x)\cos x}{\sin^2 x}$ .

Why

Naming $u$ , $v$ , $u'$ , $v'$ is the method mark. The derivative of $-\cos x$ is $+\sin x$ : the minus of the formula meets the minus in front.

Step 3

The numerator is $\sin^2 x - \cos x + \cos^2 x = 1 - \cos x$ , and $\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$ . On $(0, \pi)$ , $1 - \cos x \ne 0$ , so $f'(x) = \frac{1}{1 + \cos x}$ .

Why

The identity used twice, then a factor cancelled only because it does not vanish on the domain: say so, it is part of the answer.

Step 4

At $\frac{\pi}{2}$ : $f\left(\frac{\pi}{2}\right) = \frac{1 - 0}{1} = 1$ and $f'\left(\frac{\pi}{2}\right) = \frac{1}{1 + 0} = 1$ . Tangent: $y = 1 + \left(x - \frac{\pi}{2}\right) = x + 1 - \frac{\pi}{2}$ .

Why

A tangent line needs the point AND the slope, both exact; $\frac{\pi}{2}$ stays in the equation.

Step 5

Check by a second road: multiplying by the conjugate, $f(x) = \frac{1 - \cos^2 x}{\sin x(1 + \cos x)} = \frac{\sin x}{1 + \cos x}$ , whose derivative is $\frac{\cos x(1 + \cos x) + \sin^2 x}{(1 + \cos x)^2} = \frac{1}{1 + \cos x}$ . Same result.

Why

Two independent computations that agree are worth more than one computation checked twice. The simpler form also confirms the limit $0$ at once.

The conclusion, written out

“ $\lim_{x\to 0^+} f(x) = 0$ ; by the quotient rule and $\sin^2 x + \cos^2 x = 1$ , $f'(x) = \frac{1}{1 + \cos x}$ on $(0, \pi)$ ; the tangent at $x = \frac{\pi}{2}$ is $y = x + 1 - \frac{\pi}{2}$ .”

The classic mistake on this problem: Writing $u' = -\sin x$ for $u = 1 - \cos x$ , which gives the numerator $-\sin^2 x - \cos x + \cos^2 x$ and nothing simplifies; or cancelling $1 - \cos x$ without saying it is nonzero on $(0, \pi)$ .

Learn by heart

• $\frac{\sin\theta}{\theta} \to 1$ and $\frac{1 - \cos\theta}{\theta} \to 0$ as $\theta \to 0$ , in RADIANS; $\frac{1 - \cos\theta}{\theta^2} \to \frac{1}{2}$ .
• Same box: $\frac{\sin\square}{\square} \to 1$ only when the SAME $\square$ tends to $0$ ; $\frac{\sin ax}{x} \to a$ .
• $(\sin x)' = \cos x$ , $(\cos x)' = -\sin x$ , $(\tan x)' = \sec^2 x$ .
• $(\sec x)' = \sec x\tan x$ , $(\csc x)' = -\csc x\cot x$ , $(\cot x)' = -\csc^2 x$ : the co-functions take a minus.
• $\frac{d^n}{dx^n}\sin x$ depends on the remainder of $n$ divided by $4$ : $\sin, \cos, -\sin, -\cos$ .
• $f'(x) = 0$ : FACTOR, never divide by $\sin x$ or $\cos x$ without checking.

Frequently asked questions

Why must x be in radians to differentiate sin x?

Because the formula comes from the limit of sin h over h as h goes to zero, and that limit equals 1 only in radians: its proof uses the area of a circular sector, one half of r squared times the angle, which is a radian formula. In degrees the same limit is pi over 180, and the derivative of the sine of x degrees is pi over 180 times its cosine.

How do I remember the signs of the six trig derivatives?

The three co-functions, cosine, cotangent and cosecant, have a minus sign in their derivatives, and the three others do not. Each co-derivative is also the mirror of its partner: swap every function with its co-function and add the minus. Tangent gives secant squared, so cotangent gives minus cosecant squared; secant gives secant tangent, so cosecant gives minus cosecant cotangent.

How do I compute the limit of sin 5x over 3x as x goes to 0?

Make the same expression appear under the sine and in the denominator. Write the quotient as five thirds times sin 5x over 5x. Since 5x goes to zero, sin 5x over 5x goes to 1, and the limit is five thirds. Answering 1 directly is the most common mistake: the rule needs the same box in both places.

Can I differentiate sin 2x before learning the chain rule?

Yes. Write sin 2x as 2 sin x cos x and use the product rule: the derivative is 2 times cos squared x minus sin squared x, which is 2 cos 2x by the double angle formula. The factor 2 is the one students drop when they answer cos 2x. The chain rule of the next chapter gives the same result mechanically.

What is the 99th derivative of sin x?

The derivatives of sine repeat every four steps: cosine, minus sine, minus cosine, then sine again. Divide 99 by 4: the remainder is 3, so the 99th derivative equals the third one, minus cosine of x. Reasoning only on whether 99 is odd tells you it is a cosine, but not its sign.

Practise it

Corrected exercises: Derivatives of trigonometric functions, MATH 140 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Differentiation rules Next sheet The chain rule

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math140-trigonometric-derivatives. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: derivatives of trigonometric functions (MATH 140)

The essentials

One limit, three conditions

The six derivatives, and where each sign comes from

The rules in table form

The six derivatives, each with a slope you can check

The mistakes that cost marks

1. Using the standard limit with the wrong box

2. Answering 1 to one minus cosine over x

3. Replacing sin x and tan x by x inside a difference

4. Losing the minus sign of a co-function

5. Differentiating a quotient as the quotient of the derivatives

6. Differentiating a product as the product of the derivatives

7. Dividing by sin x when solving f'(x) = 0

8. Differentiating with the angle in degrees

9. Reading a high derivative on parity

Which method to choose

Which tool for a trigonometric limit, by the FORM of the expression

How the answer is expected to be written

Proving the derivative of cosine from the definition

Check before you hand in

The typical problem, taken apart

A quotient that collapses: limit, derivative and tangent line

Learn by heart

Frequently asked questions

Why must x be in radians to differentiate sin x?

How do I remember the signs of the six trig derivatives?

How do I compute the limit of sin 5x over 3x as x goes to 0?

Can I differentiate sin 2x before learning the chain rule?

What is the 99th derivative of sin x?

Corrected exercises: Derivatives of trigonometric functions, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?