MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: power, exponential, product and quotient rules (MATH 140)

Q: When should I use the quotient rule and when should I rewrite first?

Rewrite first when the denominator is a single power of x: divide each term of the numerator by it and use the power rule. When the numerator is a constant, use the reciprocal rule. When top and bottom have the same degree, dividing first often gives a constant plus a simpler fraction. Keep the full quotient rule for a genuine quotient of two functions.

Q: Can I use the power rule to differentiate e to the x?

No. The power rule is for a variable base raised to a constant exponent, like x to the fifth. In e to the x the base is a constant and the exponent varies, so it is an exponential function, whose derivative is e to the x itself. The power rule would give x times e to the x minus one, which is zero at x equal to zero, where e to the x actually rises with slope one.

Q: How do I find a tangent line through a point that is not on the curve?

Call a the unknown x-coordinate of the point of tangency and write the tangent line at a, keeping a as a letter. Then substitute the coordinates of the given point into that equation. You get an equation in a alone; each solution gives one tangent. Never use the slope of the curve at the x-coordinate of the given point, since that point is not where the line touches.

Q: How do I know if a particle is speeding up or slowing down?

Compute the velocity, the derivative of the position, and the acceleration, the derivative of the velocity. Put the signs of both on one sign chart. On each interval where they have the same sign the particle speeds up, and where they have opposite signs it slows down. The sign of the acceleration alone is not enough when the particle moves backward.

Q: What is the slope of the normal line when the tangent is horizontal?

It has no slope, because the normal is vertical. When the derivative at a is zero, the tangent is the horizontal line y equals f of a, and the line perpendicular to it through the point is the vertical line x equals a. The formula minus one over the derivative cannot produce it, but the normal line exists and must be given as x equals a.

This sheet is not a summary of sections 3.1, 3.2 and 3.7 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on the differentiation rules in MATH 140 at McGill University, and which precise gesture avoids each loss.

The rules look mechanical, and that is the danger: the derivative is usually right up to one slip, a product split into two derivatives, a minus sign lost in the quotient rule, a point substituted too early. Every value below is exact and done by hand, as on the exam, and every trap comes with the line that earns the mark.

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 rule applies to a FORM, so the marks are won before and after differentiating: rewrite roots and fractions as powers first, never differentiate a product or a quotient factor by factor, and substitute the point only once $f'$ is written as a 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

Power rule
Derivative of the exponential function
Sum and constant multiple rules
Product rule
Quotient rule
Higher-order derivatives
Tangent line
Normal line
Horizontal tangent
Position, velocity and acceleration

The essentials

Each rule has its form: power, exponential, constant

• Power rule: $(x^n)' = nx^{n-1}$ for EVERY real $n$ , but only for a VARIABLE base and a CONSTANT exponent. Rewrite first: $\sqrt[3]{x} = x^{1/3}$ , $\frac{1}{\sqrt x} = x^{-1/2}$ , $\frac{5}{x^2} = 5x^{-2}$ .
• Exponential: $(e^x)' = e^x$ . Constant base, variable exponent: the power rule does not apply, and $xe^{x-1}$ is wrong.
• Constants: $(\pi^3)' = 0$ , $(e^2)' = 0$ , $(\sqrt 7)' = 0$ . A number that does not contain $x$ has a horizontal graph.
• Without the chain rule, rewrite with the laws of exponents: $e^{x+2} = e^2e^x$ , $e^{-x} = \frac{1}{e^x}$ , $e^{2x} = e^xe^x$ .
• Sum and constant multiple: differentiate term by term, constants in front stay in front.

y = e^x

the slope of the tangent equals the height of the point: slope

1

(0, 1)

, slope

e

(1, e)

. No power rule gives that.

A marker reads the first line: when it rewrites $\frac{8}{\sqrt x}$ as $8x^{-1/2}$ , the method mark is already secured, whatever happens to the arithmetic afterwards.

Products, quotients, tangents: the formulas with their conditions

• Product: $(fg)' = f'g + fg'$ . Quotient: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$ , derivative of the TOP first. Reciprocal: $\left(\frac{c}{g}\right)' = -\frac{cg'}{g^2}$ .
• Second derivative of a product: $(fg)'' = f''g + 2f'g' + fg''$ , coefficients $1, 2, 1$ .
• Tangent at $a$ : $y = f(a) + f'(a)(x - a)$ , with $f'(a)$ a NUMBER. Normal: slope $-\frac{1}{f'(a)}$ ; if $f'(a) = 0$ , the normal is the vertical line $x = a$ .
• Tangent through a point $P$ off the curve: the point of tangency $a$ is UNKNOWN, write the tangent at $a$ and require that it passes through $P$ .
• Motion: $v = s'$ , $a = s''$ . Speeding up when $v$ and $a$ have the SAME sign. Distance: split at the zeros of $v$ .

The product and quotient rules need $f$ and $g$ differentiable AT the point. At a corner of one factor, the rule cannot be quoted: look at the function itself.

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 rules that exist, and the ones that do not

Read a line as: an expression of the first column, under the condition of the second, has the derivative of the third. The red lines are rules students write that do not exist: each is refuted by two numbers in its example.

Expression	Condition or tempting rule	Derivative
$x^n$	$n$ any real	$nx^{n-1}$
Example: $(x^{-2})' = -2x^{-3}$ , and $(\sqrt x)' = \frac{1}{2\sqrt x}$ equals $\frac{1}{4}$ at $x = 4$ .
$e^x$	base $e$ , exponent $x$	$e^x$
Example: Slope $1$ at $x = 0$ and slope $e$ at $x = 1$ ; $(e^{x+2})' = e^{x+2}$ .
$c$	no $x$ inside	$0$
Example: $(\pi^3)' = 0$ , $(e^2)' = 0$ , and $(x^2 + \pi^2)' = 2x$ .
$(fg)'$	$f'g'$	not a rule rule that does not exist
Example: For $x \cdot x$ : $f'g' = 1$ , but $(x^2)' = 2x = 6$ at $x = 3$ . What to do: Write $f'g + fg'$ , naming $f$ and $g$ first.
$\left(\frac{f}{g}\right)'$	$\frac{f'}{g'}$	not a rule rule that does not exist
Example: For $\frac{5}{x^2 + 4}$ : $\frac{0}{2x} = 0$ , but the true derivative is $-\frac{2}{5}$ at $x = 1$ . What to do: Write $\frac{f'g - fg'}{g^2}$ , or the reciprocal rule when the top is a constant.
$(e^x)'$	power rule $xe^{x-1}$	not a rule rule that does not exist
Example: $xe^{x-1}$ is $0$ at $x = 0$ , while $e^x$ rises there with slope $1$ . What to do: $(e^x)' = e^x$ : the power rule needs a constant exponent.

The three black lines and the product and quotient rules are the whole chapter: every derivative of the series is a combination of them, after rewriting.

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. Differentiating a product factor by factor

the whole derivative, and every question built on it

What not to write

“ $(x^2e^x)' = 2x \cdot e^x = 2xe^x$ .”

What to write

“Product rule with $f = x^2$ and $g = e^x$ : $(x^2e^x)' = 2xe^x + x^2e^x = x(x + 2)e^x$ .”

Why: A product changes when EITHER factor changes, which is where the two terms come from. At $x = 1$ the wrong answer gives $2e$ and the true slope is $3e$ .

2. Differentiating a quotient as the quotient of the derivatives

2 marks

What not to write

“ $\left(\frac{5}{x^2 + 4}\right)' = \frac{0}{2x} = 0$ .”

What to write

“Reciprocal rule: $\left(\frac{5}{x^2 + 4}\right)' = -\frac{5 \cdot 2x}{(x^2 + 4)^2} = -\frac{10x}{(x^2 + 4)^2}$ .”

Why: A derivative equal to $0$ everywhere would make the function constant, yet it goes from $\frac{5}{4}$ at $0$ to $1$ at $1$ . A constant numerator does not make the quotient constant.

3. Differentiating a fraction without rewriting it as a power

1 to 2 marks

What not to write

“ $\left(\frac{1}{3x^2}\right)' = \frac{1}{6x}$ .”

What to write

“ $\frac{1}{3x^2} = \frac{1}{3}x^{-2}$ , so its derivative is $\frac{1}{3} \cdot (-2)x^{-3} = -\frac{2}{3x^3}$ .”

Why: The rule $\left(\frac{1}{u}\right)' = \frac{1}{u'}$ does not exist. The function decreases for $x > 0$ , so a positive derivative there is impossible: rewrite as a power, then apply the power rule.

4. Treating e^(-x) like e^x

1 mark, and every sign that follows

What not to write

“ $e^x$ is its own derivative, so $(e^{-x})' = e^{-x}$ .”

What to write

“ $e^{-x} = \frac{1}{e^x}$ , so by the reciprocal rule $(e^{-x})' = -\frac{e^x}{(e^x)^2} = -e^{-x}$ .”

Why: $(e^x)' = e^x$ is about the exponent $x$ exactly. $e^{-x}$ decreases on the whole line, so its derivative is negative: rewrite with the laws of exponents before any rule.

5. Swapping the terms of the quotient rule

2 marks: every value of the derivative has the wrong sign

What not to write

“ $\left(\frac{3x - 1}{x^2 + 2}\right)' = \frac{(3x - 1)(2x) - 3(x^2 + 2)}{(x^2 + 2)^2}$ .”

What to write

“ $\left(\frac{3x - 1}{x^2 + 2}\right)' = \frac{3(x^2 + 2) - (3x - 1)(2x)}{(x^2 + 2)^2} = \frac{-3x^2 + 2x + 6}{(x^2 + 2)^2}$ .”

Why: The numerator is $f'g - fg'$ , derivative of the TOP first. At $x = 0$ the right answer is $\frac{3}{2}$ and the swapped one $-\frac{3}{2}$ .

6. Leaving f'(x) unevaluated in the tangent line

the whole question: the answer is not a line

What not to write

“Tangent to $y = x^2$ at $x = 1$ : $y = 1 + 2x(x - 1)$ .”

What to write

“ $f'(1) = 2$ , so the tangent is $y = 1 + 2(x - 1) = 2x - 1$ .”

The true tangent

y = 2x - 1

touches

y = x^2

(1, 1)

; the red dashed curve, with

f'(x)

unevaluated, is a second parabola, not a line.

Why: The slope of a line is a NUMBER. With $f'(x)$ left as a function, the equation $y = 2x^2 - 2x + 1$ is a parabola: differentiate, THEN substitute $a$ .

7. Substituting the point before differentiating

the whole question

What not to write

“ $f(x) = x^3$ , $f(2) = 8$ , so $f'(2) = (8)' = 0$ .”

What to write

“ $f'(x) = 3x^2$ , so $f'(2) = 12$ .”

Why: $f(2)$ is a number, and every number has derivative $0$ . The derivative at $2$ is the value at $2$ of the FUNCTION $f'$ : differentiate first, substitute last.

8. Using the slope at the abscissa of a point that is not on the curve

the whole question

What not to write

“Tangent to $y = x^2$ through $P(2, 3)$ : slope $f'(2) = 4$ , so $y = 3 + 4(x - 2) = 4x - 5$ .”

What to write

“Tangent at an unknown $a$ : $y = 2ax - a^2$ . Through $P$ : $3 = 4a - a^2$ , so $a = 1$ or $a = 3$ , giving $y = 2x - 1$ and $y = 6x - 9$ .”

From

P(2, 3)

two tangents touch

y = x^2

, at

(1, 1)

and

(3, 9)

; the red dashed line of slope

f'(2) = 4

misses the curve entirely.

Why: $P$ is not on the curve, since $2^2 = 4 \ne 3$ , so $f'(2)$ is the slope at $(2, 4)$ , not at the point of tangency. The line $4x - 5$ never meets the parabola: $x^2 - 4x + 5$ has discriminant $-4$ .

9. The normal line with the wrong slope, or no normal at all

1 to 2 marks per normal

What not to write

“At $(1, -6)$ the tangent has slope $-12$ , so the normal has slope $12$ . At $(-1, 10)$ the slope is $0$ , so there is no normal.”

What to write

“The normal has slope $-\frac{1}{-12} = \frac{1}{12}$ . At $(-1, 10)$ the tangent is horizontal, so the normal is the vertical line $x = -1$ .”

Why: Perpendicular slopes multiply to $-1$ : opposite AND reciprocal. A horizontal tangent has a vertical normal, which exists but has no slope.

10. Deciding speeding up from the sign of the acceleration alone

2 marks

What not to write

“For $s(t) = t^3 - 9t^2 + 24t$ , $a(t) = 6t - 18 > 0$ on $(3, 4)$ , so the particle speeds up there.”

What to write

“On $(3, 4)$ , $v(t) = 3(t - 2)(t - 4) < 0$ and $a(t) > 0$ : opposite signs, so the particle slows down.”

Why: The speed is $|v|$ . A positive acceleration on a particle moving backward reduces $|v|$ : compare the SIGNS of $v$ and $a$ on a common sign chart.

Which method to choose

Which rule, by the FORM of the function

Look at the shape of the expression before differentiating: the form picks the rule, and often a rewriting comes first

If roots of $x$ , or $x$ alone in a denominator → rewrite as powers $x^{p/q}$ , $x^{-n}$ , then the power rule term by term

Example: $\frac{8}{\sqrt x} = 8x^{-1/2}$ , derivative $-4x^{-3/2}$
If a quotient whose denominator is a single power of $x$ → divide term by term, then the power rule

Example: $\frac{x^3 - 4x + 1}{x^2} = x - 4x^{-1} + x^{-2}$
If a constant over a function, $\frac{c}{g}$ → reciprocal rule $-\frac{cg'}{g^2}$

Example: $\left(\frac{5}{x^2 + 4}\right)' = -\frac{10x}{(x^2 + 4)^2}$
If $e$ raised to anything other than $x$ → laws of exponents first: $e^2e^x$ , $\frac{1}{e^x}$ , $e^xe^x$

Example: $(e^{2x})' = (e^xe^x)' = 2e^{2x}$
If a product of two functions of $x$ → product rule; expand instead only for two short polynomials

Example: $(x^2e^x)' = x(x + 2)e^x$
If a quotient of two functions, same degree top and bottom → divide first to get a constant plus a simpler fraction, or the quotient rule

Example: $\frac{x^2 + 1}{x^2 - 1} = 1 + \frac{2}{x^2 - 1}$ , derivative $-\frac{4x}{(x^2 - 1)^2}$
If functions known only by a table or a graph → write the rule with letters, then substitute the four values at the point

Example: $(fg)'(2) = f'(2)g(2) + f(2)g'(2) = 2 + 12 = 14$

A composite such as $(x^2 + 1)^5$ or $\sqrt{x^2 + 1}$ is for the chain rule, two chapters later. Here, if no rewriting removes the composition, the function is not yet in reach.

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.

A tangent line through a point that is not on the curve

When to use it: The statement asks for the tangents to $y = f(x)$ that pass through a given point $P$ , and $P$ does not satisfy the equation of the curve

1 Check that $P$ is not on the curve, and SAY it: this is what forbids using the abscissa of $P$ .
2 Call $a$ the abscissa of the unknown point of tangency, and write the tangent there with $a$ as a letter: $y = f(a) + f'(a)(x - a)$ .
3 Substitute the coordinates of $P$ for $x$ and $y$ : this gives an equation in $a$ only.
4 Solve for $a$ , then write each tangent with its slope as a number, and its point of tangency.
5 Check that each line passes through $P$ .

Concluding sentence

“ $P(1, -3)$ is not on the parabola, since $1^2 \ne -3$ . The tangent at the point of abscissa $a$ is $y = 2ax - a^2$ ; it passes through $P$ when $a^2 - 2a - 3 = 0$ , so $a = 3$ or $a = -1$ . The tangents are $y = 6x - 9$ and $y = -2x - 1$ .”

The trap: Writing the tangent with the slope $f'$ at the abscissa of $P$ : the line then usually misses the curve.

Marking: Typically 2 marks for the tangent at a general point, 3 for the equation in a, 3 for its solutions and the lines, 2 for the checks.

Speeding up or slowing down

When to use it: A position $s(t)$ is given and the question asks when the particle speeds up or slows down

1 Compute $v = s'$ and $a = v' = s''$ , and factor both.
2 Find the zeros of $v$ and of $a$ and place them all on one time axis.
3 Fill a sign chart with one row for $v$ and one for $a$ .
4 On each interval, same signs means speeding up, opposite signs means slowing down; write the intervals.

Concluding sentence

“On $(2, 3)$ , $v < 0$ and $a < 0$ have the same sign, so the particle speeds up; on $(3, 4)$ , $v < 0$ and $a > 0$ have opposite signs, so it slows down.”

The trap: Answering from the sign of the acceleration alone, which is correct only while the particle moves in the positive direction.

Marking: Typically 2 marks for v and a, 3 for the sign chart, 3 for the intervals, 2 for the sentence.

Check before you hand in

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

Expand to check a product

When both factors are polynomials, expand the product and differentiate term by term: the two answers must agree, and the degree of the derivative must be one less than the degree of the product.

$(x^2 + 1)(x^3 - 2x) = x^5 - x^3 - 2x$ gives $5x^4 - 3x^2 - 2$ ; the factor-by-factor answer $2x(3x^2 - 2)$ has degree $3$ and fails.
The sign matches the graph

A function that decreases on an interval has a negative derivative there. Test one point of the interval in your answer.

$\frac{1}{3x^2}$ decreases for $x > 0$ : the answer $\frac{1}{6x} > 0$ is impossible, $-\frac{2}{3x^3} < 0$ is consistent.
The tangent passes through its point

Plug the abscissa of the point of tangency into your tangent line: you must get f(a). Then check that the slope is a number, not an expression in x.

$y = -12x + 6$ at $x = 1$ gives $-6 = f(1)$ : consistent. $y = 2x - 5$ through $P(1, -3)$ passes through $P$ but never meets $y = x^2$ .
Distance at least the displacement

The total distance travelled is never smaller than the absolute value of the displacement, and equals it only if the velocity never changes sign.

Displacement $36$ m, distance $44$ m, because $v$ changes sign twice; a distance of $36$ m would mean no turning back.

The typical problem, taken apart

Horizontal tangents, a tangent and a normal on a rational function

Let $f(x) = \frac{x^2 + 3}{x + 1}$ , for $x \ne -1$ . Find the points where the tangent is horizontal, then the tangent and the normal lines at the point of abscissa $0$ .

No calculator. Every rule must be named, as on a MATH 140 final.

The graph has two branches on either side of

x = -1

: the flat points to find are the top of the left branch and the bottom of the right one.

Step 1

Quotient rule with top $x^2 + 3$ (derivative $2x$ ) and bottom $x + 1$ (derivative $1$ ): $f'(x) = \frac{2x(x + 1) - (x^2 + 3) \cdot 1}{(x + 1)^2}$ .

Why

Naming the rule and the two functions earns the method mark before any algebra. The denominator $(x + 1)^2$ is left squared, never expanded.

Step 2

Numerator: $2x^2 + 2x - x^2 - 3 = x^2 + 2x - 3 = (x + 3)(x - 1)$ , so $f'(x) = \frac{(x + 3)(x - 1)}{(x + 1)^2}$ .

Why

The factored form is the useful one: it gives the zeros of $f'$ at once, and its sign if a later question asks for it.

Step 3

$f'(x) = 0$ when $x = -3$ or $x = 1$ , both in the domain. $f(-3) = \frac{12}{-2} = -6$ and $f(1) = \frac{4}{2} = 2$ : horizontal tangents at $(-3, -6)$ and $(1, 2)$ .

Why

A fraction is zero when its NUMERATOR is zero and its denominator is not. The question asks for points, so both coordinates are expected.

Step 4

At $x = 0$ : $f(0) = 3$ and $f'(0) = \frac{3 \cdot (-1)}{1} = -3$ . Tangent: $y = 3 - 3x$ . Normal: slope $-\frac{1}{-3} = \frac{1}{3}$ , so $y = 3 + \frac{1}{3}x$ .

Why

The slope is computed from the factored $f'$ , AFTER differentiating. The normal takes the opposite of the reciprocal, both changes at once.

Step 5

Checks: $(-3) \cdot \frac{1}{3} = -1$ , so the lines are perpendicular; both pass through $(0, 3)$ ; and the flat points match the figure, a top at $-6$ on the left and a bottom at $2$ on the right.

Why

Three checks in ten seconds: the perpendicularity, the point, and the picture. They catch a swapped quotient rule, which would put the flat points at the wrong places.

The conclusion, written out

“The tangent is horizontal at $(-3, -6)$ and at $(1, 2)$ . At $(0, 3)$ the tangent is $y = -3x + 3$ and the normal is $y = \frac{1}{3}x + 3$ .”

The classic mistake on this problem: Writing the numerator as $(x^2 + 3) - 2x(x + 1)$ , the terms swapped: the zeros come out the same, which hides the error, but the tangent at $0$ gets slope $+3$ and the normal slope $-\frac{1}{3}$ .

Learn by heart

• $(x^n)' = nx^{n-1}$ for every real $n$ , after rewriting roots and fractions as powers.
• $(e^x)' = e^x$ ; $(c)' = 0$ for every constant, $\pi^3$ and $e^2$ included.
• $(fg)' = f'g + fg'$ , never $f'g'$ .
• $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$ , top first; $\left(\frac{c}{g}\right)' = -\frac{cg'}{g^2}$ .
• Tangent: $y = f(a) + f'(a)(x - a)$ , with $f'(a)$ a NUMBER. Normal: $-\frac{1}{f'(a)}$ , or $x = a$ when $f'(a) = 0$ .
• Point off the curve: unknown point of tangency $a$ , then the condition through $P$ .
• Speeding up: $v$ and $a$ of the SAME sign. Distance: split at the zeros of $v$ .

Frequently asked questions

When should I use the quotient rule and when should I rewrite first?

Rewrite first when the denominator is a single power of x: divide each term of the numerator by it and use the power rule. When the numerator is a constant, use the reciprocal rule. When top and bottom have the same degree, dividing first often gives a constant plus a simpler fraction. Keep the full quotient rule for a genuine quotient of two functions.

Can I use the power rule to differentiate e to the x?

No. The power rule is for a variable base raised to a constant exponent, like x to the fifth. In e to the x the base is a constant and the exponent varies, so it is an exponential function, whose derivative is e to the x itself. The power rule would give x times e to the x minus one, which is zero at x equal to zero, where e to the x actually rises with slope one.

How do I find a tangent line through a point that is not on the curve?

Call a the unknown x-coordinate of the point of tangency and write the tangent line at a, keeping a as a letter. Then substitute the coordinates of the given point into that equation. You get an equation in a alone; each solution gives one tangent. Never use the slope of the curve at the x-coordinate of the given point, since that point is not where the line touches.

How do I know if a particle is speeding up or slowing down?

Compute the velocity, the derivative of the position, and the acceleration, the derivative of the velocity. Put the signs of both on one sign chart. On each interval where they have the same sign the particle speeds up, and where they have opposite signs it slows down. The sign of the acceleration alone is not enough when the particle moves backward.

What is the slope of the normal line when the tangent is horizontal?

It has no slope, because the normal is vertical. When the derivative at a is zero, the tangent is the horizontal line y equals f of a, and the line perpendicular to it through the point is the vertical line x equals a. The formula minus one over the derivative cannot produce it, but the normal line exists and must be given as x equals a.

Practise it

Corrected exercises: Differentiation rules, 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 The derivative as a limit Next sheet Derivatives of trigonometric functions

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

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Revision sheet: power, exponential, product and quotient rules (MATH 140)

The essentials

Each rule has its form: power, exponential, constant

Products, quotients, tangents: the formulas with their conditions

The rules in table form

The rules that exist, and the ones that do not

The mistakes that cost marks

1. Differentiating a product factor by factor

2. Differentiating a quotient as the quotient of the derivatives

3. Differentiating a fraction without rewriting it as a power

4. Treating e^(-x) like e^x

5. Swapping the terms of the quotient rule

6. Leaving f'(x) unevaluated in the tangent line

7. Substituting the point before differentiating

8. Using the slope at the abscissa of a point that is not on the curve

9. The normal line with the wrong slope, or no normal at all

10. Deciding speeding up from the sign of the acceleration alone

Which method to choose

Which rule, by the FORM of the function

How the answer is expected to be written

A tangent line through a point that is not on the curve

Speeding up or slowing down

Check before you hand in

The typical problem, taken apart

Horizontal tangents, a tangent and a normal on a rational function

Learn by heart

Frequently asked questions

When should I use the quotient rule and when should I rewrite first?

Can I use the power rule to differentiate e to the x?

How do I find a tangent line through a point that is not on the curve?

How do I know if a particle is speeding up or slowing down?

What is the slope of the normal line when the tangent is horizontal?

Corrected exercises: Differentiation rules, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?