MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: derivatives of trigonometric functions (MATH 203)

Q: What are the derivatives of the six trigonometric functions?

Sine gives cosine, tangent gives secant squared, secant gives secant times tangent. The three co-functions take a minus sign: cosine gives minus sine, cotangent gives minus cosecant squared, cosecant gives minus cosecant times cotangent. All six assume the angle is in radians.

Q: Should I simplify a trig expression before or after differentiating?

Both, and the form decides. If the function rewrites as one of the six, or a complex fraction collapses with sin squared plus cos squared equals one, simplify first: the derivative then takes one line. If it does not simplify, use the quotient rule, then expand the numerator and look for the same identity, which usually reduces it to a short answer.

Q: Why does my calculator give the wrong slope for sin x?

It is almost certainly in degree mode. The formula saying the derivative of sin x is cos x holds only in radians, so cos 2 must be computed in radians: about minus 0.4161, not 0.9994. Test the mode before the exam by computing cos of pi, which must display minus 1.

Q: How do I find all the horizontal tangents of a trig function between 0 and 2 pi?

Set the derivative equal to zero, then factor it instead of dividing, using sin squared equals one minus cos squared to get a single function if needed. Solve each factor on the unit circle: a value strictly between minus one and one gives two angles per turn, while the calculator returns only one. Give both coordinates of each point.

Q: How do I differentiate sec squared x before learning the chain rule?

Write it as the product sec x times sec x and use the product rule: sec x tan x times sec x plus sec x times sec x tan x, which is 2 sec squared x tan x. Applying the power rule and writing 2 sec x is wrong, because the power rule is about a power of x, not a power of a function.

This sheet is not a summary of section 3.5 of Thomas' Calculus: you already have the lecture notes. It answers one question only, what makes students lose marks on the derivatives of trigonometric functions in MATH 203 at Concordia University, and which precise gesture avoids each loss.

The six formulas fit on one line and are rarely the problem. The marks are lost in the algebra around them: an identity not seen before differentiating, a numerator left unsimplified after, an equation $f'(x) = 0$ divided instead of factored, and a calculator that returns one angle, sometimes in the wrong mode. The calculator is allowed in MATH 203; it never replaces the unit circle.

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

Differentiating a trigonometric function takes one line; the marks go to the ALGEBRA around it: the identity $\sin^2 x + \cos^2 x = 1$ used BEFORE (rewrite, one fraction, cancel a factor), AFTER (collapse the numerator) and IN $f'(x) = 0$ (one function, factor, every quadrant), plus the three minus signs of the co-functions.

This chapter is part of MATH 203, Calculus I (Concordia)

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

Derivatives of trigonometric functions
Derivative of sine and cosine
Reciprocal trigonometric functions
Pythagorean identities
Product rule
Quotient rule
Higher-order derivatives
Horizontal tangent line
Normal line
Trigonometric equations
Radian measure
Simple harmonic motion

The essentials

Six derivatives, three minus signs

• $(\sin x)' = \cos x$ , $(\tan x)' = \sec^2 x$ , $(\sec x)' = \sec x\tan x$ .
• $(\cos x)' = -\sin x$ , $(\cot x)' = -\csc^2 x$ , $(\csc x)' = -\csc x\cot x$ : the three CO-functions take a minus.
• The pairs never cross: tangent and cotangent give SQUARES ( $\sec^2$ , $-\csc^2$ ), secant and cosecant give PRODUCTS ( $\sec\tan$ , $-\csc\cot$ ).
• All six assume $x$ in RADIANS; Thomas builds them in 3.5 from the limits of section 2.4, which are radian limits. The calculator goes in RAD mode before the exam starts.
• The derivatives of $\sin x$ and $\cos x$ repeat every four steps: $\sin \to \cos \to -\sin \to -\cos \to \sin$ .

\cot x

falls across the whole of

(0, \pi)

, so its slope

-\csc^2 x

is negative everywhere, at best

-1

\frac{\pi}{2}

: the minus sign can be SEEN.

A minus in front of a co-function meets the minus of its derivative: $(-4\csc x)' = +4\csc x\cot x$ and $(-20\cos t)' = +20\sin t$ .

One identity, three disguises, three moments

• $\sin^2 x + \cos^2 x = 1$ ; divided by $\cos^2 x$ : $\tan^2 x + 1 = \sec^2 x$ ; divided by $\sin^2 x$ : $1 + \cot^2 x = \csc^2 x$ .
• BEFORE differentiating: rewrite in $\sin$ and $\cos$ , bring the complex fraction to ONE fraction, cancel only a FACTOR. $\frac{\sec x - \cos x}{\sin x} = \frac{\sin^2 x}{\cos x\sin x} = \tan x$ .
• AFTER the quotient rule: expand the numerator and look for $\sin^2 x + \cos^2 x$ or $\tan^2 x - \sec^2 x$ . $\left(\frac{1 - \cos x}{1 + \cos x}\right)' = \frac{2\sin x}{(1 + \cos x)^2}$ .
• IN $f'(x) = 0$ : replace $\sin^2 x$ by $1 - \cos^2 x$ to get ONE function, factor the quadratic, set each factor to $0$ .
• The domain never simplifies: $\frac{\sec x}{\tan x} = \csc x$ only where $\sin x \ne 0$ AND $\cos x \ne 0$ .

The identity is chosen by the REST of the expression: a denominator in cosines calls for $\sin^2 x = 1 - \cos^2 x$ , a $\sec^2 x$ next to $\tan x$ calls for $\sec^2 x = 1 + \tan^2 x$ .

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.

Rewrite first, then differentiate: each line 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, valid where the ORIGINAL function is defined. The red line is a cancellation that does not exist.

Function	Rewritten as	Derivative
$\cot x\sec x$	$\csc x$	$-\csc x\cot x$
Example: At $\frac{\pi}{4}$ : $-\sqrt 2\cdot 1 = -\sqrt 2$ .
$\frac{1 - \cos^2 x}{\sin x}$	$\sin x$	$\cos x$
Example: At $\frac{\pi}{3}$ : $\cos\frac{\pi}{3} = \frac{1}{2}$ ; the function is undefined at $0$ and $\pi$ .
$\sin x\,(\cot x + \tan x)$	$\frac{\cos^2 x + \sin^2 x}{\cos x} = \sec x$	$\sec x\tan x$
Example: At $\frac{\pi}{3}$ : $2\cdot\sqrt 3 = 2\sqrt 3$ .
$(\sec x - 1)(\sec x + 1)$	$\sec^2 x - 1 = \tan^2 x$	$2\tan x\sec^2 x$
Example: At $\frac{\pi}{4}$ : $2\cdot 1\cdot 2 = 4$ , by the product rule on $\tan x\cdot\tan x$ .
$\frac{\sec x}{\csc x}$	$\frac{\sin x}{\cos x} = \tan x$	$\sec^2 x$
Example: At $\frac{\pi}{3}$ : $\sec^2\frac{\pi}{3} = 4$ .
$\frac{\sin x}{\sin x + \cos x}$	$\frac{1}{\cos x}$	$\sec x\tan x$ cancellation that is false
Example: At $\frac{\pi}{4}$ the function is $\frac{\sqrt 2/2}{\sqrt 2} = \frac{1}{2}$ , while $\frac{1}{\cos(\pi/4)} = \sqrt 2$ . What to do: No rewrite: the quotient rule gives the numerator $\cos x(\sin x + \cos x) - \sin x(\cos x - \sin x) = 1$ , so the derivative is $\frac{1}{(\sin x + \cos x)^2}$ .

Only a FACTOR common to the whole numerator and the whole denominator cancels; $\sin x$ is a term of $\sin x + \cos x$ , not a factor.

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. Losing the plus sign of minus times minus

1 mark, and every value or tangent line built on it

What not to write

“ $f(x) = 7\sin x - 4\csc x$ , so $f'(x) = 7\cos x - 4\csc x\cot x$ .”

What to write

“ $f'(x) = 7\cos x - 4(-\csc x\cot x) = 7\cos x + 4\csc x\cot x$ .”

Why: The minus written in front of $\csc x$ and the minus of $(\csc x)'$ make a plus. Write the bracket $-4(-\csc x\cot x)$ before simplifying.

2. Crossing the derivatives of cotangent and cosecant

1 mark per derivative

What not to write

“ $(\cot x)' = -\csc x\cot x$ .”

What to write

“ $(\cot x)' = -\csc^2 x$ , the partner of $(\tan x)' = \sec^2 x$ ; it is $(\csc x)'$ that equals $-\csc x\cot x$ .”

Why: At $\frac{\pi}{2}$ the wrong formula gives $0$ , a flat tangent, while $\cot x$ crosses the axis there with slope $-1$ , as the first figure shows.

3. Applying the power rule to a power of a trigonometric function

the whole derivative

What not to write

“ $(\sec^2 x)' = 2\sec x$ , so $(\tan x)'' = 2\sec x$ .”

What to write

“ $(\sec x\cdot\sec x)' = \sec x\tan x\cdot\sec x + \sec x\cdot\sec x\tan x = 2\sec^2 x\tan x$ .”

Why: The power rule is about $x^n$ . Until the chain rule, a power of a function is a PRODUCT. Check at $0$ : $(\tan x)''$ is odd, so it is $0$ there, not $2$ .

4. Writing the quotient rule numerator in the wrong order

the whole derivative: the sign of every value is wrong

What not to write

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

What to write

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

Why: $u'v - uv'$ : derivative of the TOP first. The swapped order gives exactly the opposite: $-1$ instead of $1$ at $x = 0$ .

5. Cancelling a term instead of a factor

the whole question, since the derivative is taken from the wrong function

What not to write

“ $\frac{\sin x + \tan x}{\sin x} = 1 + \tan x$ .”

What to write

“ $\frac{\sin x + \tan x}{\sin x} = 1 + \frac{\tan x}{\sin x} = 1 + \sec x$ , for $\sin x \ne 0$ .”

Why: Dividing a sum divides EACH term. At $\frac{\pi}{4}$ the function is $1 + \sqrt 2$ , not $2$ .

6. Computing a slope with the calculator in degree mode

every numerical answer of the question

What not to write

“The slope of $y = \sin x$ at $x = 2$ is $\cos 2 = 0.9994$ .”

What to write

“In radian mode, $\cos 2 \approx -0.4161$ : the slope is negative, as it must be just past $\frac{\pi}{2}$ .”

Why: $(\sin x)' = \cos x$ is a radian formula. Test the mode before the exam: $\cos\pi$ must show $-1$ , not $0.9985$ .

7. Keeping only the angle the calculator returns

half the points of the question

What not to write

“ $\cos x = \frac{1}{2}$ , and $\cos^{-1}(0.5) = 1.0472$ , so on $[0, 2\pi]$ the only solution is $x = \frac{\pi}{3}$ .”

What to write

“On $[0, 2\pi]$ , $\cos x = \frac{1}{2}$ at $x = \frac{\pi}{3}$ and at $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$ .”

The line

y = \frac{1}{2}

meets

y = \cos x

twice on

[0, 2\pi]

: the calculator gives the green point, the unit circle also gives the red one.

Why: $\cos^{-1}$ returns ONE angle, in $[0, \pi]$ . For $-1 < c < 1$ , the unit circle gives two per turn: for cosine, $\pm$ the angle; for sine, the angle and $\pi$ minus it.

8. Dividing an equation by a factor that can vanish

3 of the 5 solutions, often half the question

What not to write

“ $2\sin^2 x = \sin x$ , so $2\sin x = 1$ , so $x = \frac{\pi}{6}$ or $\frac{5\pi}{6}$ on $[0, 2\pi]$ .”

What to write

“ $\sin x\,(2\sin x - 1) = 0$ , so $\sin x = 0$ or $\sin x = \frac{1}{2}$ : $x = 0, \frac{\pi}{6}, \frac{5\pi}{6}, \pi, 2\pi$ .”

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

Which method to choose

Before, after, or as it is: where the identity goes, by the FORM of the function

Look at the function before choosing a rule

If a product or quotient of trig functions that is itself one of the six → rewrite in sines and cosines first, then use the table

Example: $\frac{\sec x}{\csc x} = \tan x$ , derivative $\sec^2 x$
If $\sin^2 x$ over $1 \pm \cos x$ , or $\cos^2 x$ over $1 \pm \sin x$ → Pythagoras, difference of squares, cancel the common factor

Example: $\frac{\cos^2 x}{1 + \sin x} = 1 - \sin x$ , derivative $-\cos x$
If a sum of two fractions → common denominator, then the identity in the numerator

Example: $\frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = 2\csc x$ , derivative $-2\csc x\cot x$
If a quotient that does not simplify → quotient rule in the order $u'v - uv'$ , then expand and look for $\sin^2 x + \cos^2 x$

Example: $\left(\frac{\sin x}{\sin x + \cos x}\right)' = \frac{1}{(\sin x + \cos x)^2}$
If a power of a trig function, such as $\sec^2 x$ or $\sin^3 x$ → write it as a product and use the product rule

Example: $(\sin x\cdot\sin^2 x)' = 3\sin^2 x\cos x$

Whatever the branch, the domain of the ORIGINAL function stays: a simplified form may be defined at points where the function is not.

Solving f'(x) = 0 on an interval, by the shape of the equation

Look at the equation f'(x) = 0 once the derivative is simplified

If a product equal to zero → set each factor to zero; say which factors can never vanish

Example: $\sec x\,(2\tan x - \sec x) = 0$ : only $2\sin x = 1$
If $\sin^2 x$ and $\cos x$ mixed, or $\cos^2 x$ and $\sin x$ → one function by Pythagoras, then factor the quadratic

Example: $-2\cos^2 x + \cos x + 1 = -(2\cos x + 1)(\cos x - 1)$
If $a\sin x + b\cos x = 0$ → check that $\cos x = 0$ is not a solution, then divide: $\tan x = -\frac{b}{a}$

Example: $\sin x - \sqrt 3\cos x = 0$ : $\tan x = \sqrt 3$ , $x = \frac{\pi}{3}, \frac{4\pi}{3}$
If a square equal to a constant, such as $\sec^2 x = 4$ → take $\pm$ , then keep or reject each sign for a stated reason

Example: $\cos x = \pm\frac{1}{2}$ ; on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ only $+$ , so $x = \pm\frac{\pi}{3}$
If a basic equation $\cos x = c$ or $\sin x = c$ with $-1 < c < 1$ → two solutions per turn: the calculator angle and its partner on the unit circle

Example: $\sin x = -\frac{1}{2}$ on $[0, 2\pi]$ : $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$

Never divide by $x$ , $\sin x$ or $\cos x$ without first saying why it cannot be $0$ at a solution.

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.

Finding every horizontal tangent on an interval

When to use it: Any question that asks for the points where the tangent is horizontal, or parallel to a given line, on an interval such as $[0, 2\pi]$

1 Differentiate, naming the rule, and simplify with the identity the expression calls for.
2 Write $f'(x)$ as a PRODUCT: factor a common function, or turn it into one function by Pythagoras and factor the quadratic.
3 Set each factor to $0$ , and state which factors can never vanish, with the reason ( $\sec x \ne 0$ , $e^x > 0$ ).
4 Solve each basic equation on the interval with the unit circle: every quadrant, the endpoints if the interval is closed, and reject the points outside the domain.
5 Give each point with BOTH coordinates, in exact form.

Concluding sentence

“On $[0, 2\pi]$ , $h'(x) = -(2\sin x - 1)(\sin x + 1) = 0$ exactly when $\sin x = \frac{1}{2}$ or $\sin x = -1$ , that is at $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$ ; the horizontal tangents are at $\left(\frac{\pi}{6}, \frac{3\sqrt 3}{4}\right)$ , $\left(\frac{5\pi}{6}, -\frac{3\sqrt 3}{4}\right)$ and $\left(\frac{3\pi}{2}, 0\right)$ .”

The trap: For $h(x) = \cos x\,(1 + \sin x)$ , the factor $\sin x + 1$ vanishes only once per turn, at $\frac{3\pi}{2}$ : it is the solution students forget, or lose by dividing.

Marking: Typically 3 marks for the simplified derivative, 3 for the factored form, 2 for all the solutions, 2 for the coordinates.

Check before you hand in

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

The radian test

Before the first numerical answer, compute cos of pi on your calculator. It must display -1. If it displays 0.9985, the calculator is in degrees and every slope will be wrong.

In degrees the calculator gives $\cos 2 = 0.9994$ , a positive slope for $\sin x$ at $x = 2$ , where the graph is plainly coming down.
The slope at a special angle

Evaluate your derivative at 0, pi over 4 or pi over 2 and compare with what the graph says there: flat at a peak, slope 1 for sine and tangent at the origin.

$(\cot x)' = -\csc x\cot x$ fails at $\frac{\pi}{2}$ : it gives $0$ , while $\cot x$ crosses the axis there with slope $-1$ .
Parity

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

$\sec x$ is even, so $(\sec x)' = \sec x\tan x$ must be odd, and it is; $2\sec x$ as a second derivative of the odd $\tan x$ is even, so it is wrong.
Two forms, one angle

After simplifying with an identity, evaluate the raw form and the simplified form at one convenient angle. They must agree.

$-\csc x\cot^2 x - \csc^3 x$ and $-\csc x\,(2\csc^2 x - 1)$ both give $-14$ at $\frac{\pi}{6}$ .
Count on the unit circle

For cos x = c or sin x = c with c strictly between -1 and 1, expect exactly two solutions per turn. One solution means one is missing.

$\cos^2 x = \frac{1}{2}$ on $[0, 2\pi]$ gives four angles, $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ , not the single $0.7854$ of the calculator.

The typical problem, taken apart

One horizontal tangent, found by factoring, and one tangent line

Let $f(x) = \sec x - 2\cos x$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ . Find every point where the tangent is horizontal, then the tangent line at $x = \frac{\pi}{3}$ .

Exact values; name every rule and every identity, as on a MATH 203 final.

The graph is symmetric about the

y

-axis and has exactly one flat point, the bottom

(0, -1)

; the dashed lines

x = \pm\frac{\pi}{2}

are where

\sec x

is undefined.

Step 1

Sum and constant multiple rules: $f'(x) = \sec x\tan x - 2(-\sin x) = \sec x\tan x + 2\sin x$ .

Why

The minus in front of $2\cos x$ meets the minus of $(\cos x)'$ : this plus is the first mark of the question.

Step 2

Rewrite $\sec x\tan x = \frac{\sin x}{\cos^2 x} = \sin x\sec^2 x$ and factor: $f'(x) = \sin x\left(\sec^2 x + 2\right)$ .

Why

Rewriting in sines and cosines makes the common factor $\sin x$ visible; a product is what an equation $f' = 0$ needs.

Step 3

$\sec^2 x + 2 \ge 3 > 0$ , so $f'(x) = 0 \iff \sin x = 0 \iff x = 0$ on the interval. $f(0) = 1 - 2 = -1$ : horizontal tangent at $(0, -1)$ .

Why

Saying why the second factor never vanishes is the step that proves there is only one point; without it the answer is a guess.

Step 4

At $\frac{\pi}{3}$ : $f = 2 - 2\cdot\frac{1}{2} = 1$ and $f' = \frac{\sqrt 3}{2}(4 + 2) = 3\sqrt 3$ . Tangent: $y = 1 + 3\sqrt 3\left(x - \frac{\pi}{3}\right)$ .

Why

The factored form gives the slope with less arithmetic; $\sec^2\frac{\pi}{3} = 4$ is read off $\cos\frac{\pi}{3} = \frac{1}{2}$ .

Step 5

Check: from the unfactored form, $\sec\frac{\pi}{3}\tan\frac{\pi}{3} + 2\sin\frac{\pi}{3} = 2\sqrt 3 + \sqrt 3 = 3\sqrt 3$ . And $f$ is even, so $f'$ must be odd: $\sin x\left(\sec^2 x + 2\right)$ is.

Why

Two forms that agree at one angle, and a parity that matches the graph: ten seconds, and the sign errors are caught.

The conclusion, written out

“ $f'(x) = \sin x\left(\sec^2 x + 2\right)$ vanishes on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ only at $x = 0$ , since $\sec^2 x + 2 > 0$ : the only horizontal tangent is at $(0, -1)$ . The tangent at $x = \frac{\pi}{3}$ is $y = 1 + 3\sqrt 3\left(x - \frac{\pi}{3}\right)$ .”

The classic mistake on this problem: Writing $f'(x) = \sec x\tan x - 2\sin x$ , which vanishes at points that are not flat at all; or dividing $f'(x) = 0$ by $\sin x$ and concluding that there is no horizontal tangent.

Learn by heart

• $(\sin x)' = \cos x$ , $(\tan x)' = \sec^2 x$ , $(\sec x)' = \sec x\tan x$ .
• $(\cos x)' = -\sin x$ , $(\cot x)' = -\csc^2 x$ , $(\csc x)' = -\csc x\cot x$ : the co-functions take a minus.
• $\sin^2 x + \cos^2 x = 1$ , $1 + \tan^2 x = \sec^2 x$ , $1 + \cot^2 x = \csc^2 x$ .
• Quotient rule: $\frac{u'v - uv'}{v^2}$ , derivative of the TOP first.
• $\frac{d^n}{dx^n}\sin x$ and $\frac{d^n}{dx^n}\cos x$ depend on the remainder of $n$ divided by $4$ .
• $\cos x = c$ , $-1 < c < 1$ : TWO solutions per turn. Calculator in RADIANS.

Frequently asked questions

What are the derivatives of the six trigonometric functions?

Sine gives cosine, tangent gives secant squared, secant gives secant times tangent. The three co-functions take a minus sign: cosine gives minus sine, cotangent gives minus cosecant squared, cosecant gives minus cosecant times cotangent. All six assume the angle is in radians.

Should I simplify a trig expression before or after differentiating?

Both, and the form decides. If the function rewrites as one of the six, or a complex fraction collapses with sin squared plus cos squared equals one, simplify first: the derivative then takes one line. If it does not simplify, use the quotient rule, then expand the numerator and look for the same identity, which usually reduces it to a short answer.

Why does my calculator give the wrong slope for sin x?

It is almost certainly in degree mode. The formula saying the derivative of sin x is cos x holds only in radians, so cos 2 must be computed in radians: about minus 0.4161, not 0.9994. Test the mode before the exam by computing cos of pi, which must display minus 1.

How do I find all the horizontal tangents of a trig function between 0 and 2 pi?

Set the derivative equal to zero, then factor it instead of dividing, using sin squared equals one minus cos squared to get a single function if needed. Solve each factor on the unit circle: a value strictly between minus one and one gives two angles per turn, while the calculator returns only one. Give both coordinates of each point.

How do I differentiate sec squared x before learning the chain rule?

Write it as the product sec x times sec x and use the product rule: sec x tan x times sec x plus sec x times sec x tan x, which is 2 sec squared x tan x. Applying the power rule and writing 2 sec x is wrong, because the power rule is about a power of x, not a power of a function.

Practise it

Corrected exercises: Derivatives of trigonometric functions, MATH 203 at Concordia

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/math203-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 203)

The essentials

Six derivatives, three minus signs

One identity, three disguises, three moments

The rules in table form

Rewrite first, then differentiate: each line with a slope you can check

The mistakes that cost marks

1. Losing the plus sign of minus times minus

2. Crossing the derivatives of cotangent and cosecant

3. Applying the power rule to a power of a trigonometric function

4. Writing the quotient rule numerator in the wrong order

5. Cancelling a term instead of a factor

6. Computing a slope with the calculator in degree mode

7. Keeping only the angle the calculator returns

8. Dividing an equation by a factor that can vanish

Which method to choose

Before, after, or as it is: where the identity goes, by the FORM of the function

Solving f'(x) = 0 on an interval, by the shape of the equation

How the answer is expected to be written

Finding every horizontal tangent on an interval

Check before you hand in

The typical problem, taken apart

One horizontal tangent, found by factoring, and one tangent line

Learn by heart

Frequently asked questions

What are the derivatives of the six trigonometric functions?

Should I simplify a trig expression before or after differentiating?

Why does my calculator give the wrong slope for sin x?

How do I find all the horizontal tangents of a trig function between 0 and 2 pi?

How do I differentiate sec squared x before learning the chain rule?

Corrected exercises: Derivatives of trigonometric functions, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?