MATH 140 derivative definition: revision sheet

The thread of the chapter

$f'(a)$ is the LIMIT of the slope of a secant, and that slope is a $\frac{0}{0}$ form by construction: rewrite the quotient until $h$ cancels BEFORE letting $h \to 0$ , and remember that the derivative exists only when the slopes from the left and from the right are finite and EQUAL.

Topics covered

Tangent line
Secant line
Difference quotient
Derivative at a point
Derivative as a function
Instantaneous rate of change
Average rate of change
Differentiability
One-sided derivatives
Vertical tangent
Corners and cusps
Continuity and differentiability

The essentials

One definition, two forms, and a 0/0 form every time

• Slope of the secant through $(a, f(a))$ and $(a + h, f(a + h))$ : $\frac{f(a+h) - f(a)}{h}$ . The tangent is the limit of the secants as $h \to 0$ .
• $f'(a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h} = \lim_{x\to a} \frac{f(x) - f(a)}{x - a}$ . Both forms are the same limit, with $x = a + h$ .
• The numerator always tends to $f(a) - f(a) = 0$ and the denominator to $0$ : substituting gives $\frac{0}{0}$ for every function, which says nothing.
• So the quotient is REWRITTEN for $h \ne 0$ until $h$ cancels: expand a polynomial, factor $(x - a)$ , multiply a root by its conjugate, take a common denominator.
• Tangent line: $y = f(a) + f'(a)(x - a)$ . As a function, $f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$ , with $x$ held fixed during the limit.

y = x^2

, the secants from

P(1, 1)

have slopes

3

for

h = 1

and

2.5

for

h = 0.5

, that is

2 + h

: they turn toward the tangent of slope

2

Write the limit symbol on EVERY line until the last one, where $h$ disappears. A line $\frac{2h + h^2}{h} = 2 + h = 2$ without $\lim$ is false as written, and markers take a mark for it.

When the derivative does not exist

• $f$ is differentiable at $a$ when $\lim_{h\to 0^-}$ and $\lim_{h\to 0^+}$ of the quotient exist, are FINITE, and are EQUAL.
• Corner: two different finite slopes. $|x|$ at $0$ : $-1$ and $1$ ; $|x^2 - 4|$ at $2$ : $-4$ and $4$ .
• Vertical tangent: the quotient tends to $+\infty$ (or $-\infty$ ) on both sides. $\sqrt[3]{x}$ at $0$ : $\frac{1}{h^{2/3}} \to +\infty$ .
• Cusp: $-\infty$ on one side, $+\infty$ on the other. $\sqrt{|x|}$ at $0$ .
• Differentiable implies continuous, so every jump or hole kills the derivative. The converse is false: all three graphs of the figure are continuous.

Three continuous graphs, three ways to have no derivative: a corner at

2

, a vertical tangent at

7

, a cusp at

12

A formula with an absolute value is not automatically non-differentiable: $x|x|$ has quotient $|h| \to 0$ at $0$ , so it is differentiable there. Compute the quotient, never judge on the formula.

What f'(a) means, in words and in units

• $f'(a)$ is a RATE at an instant, in units of $f$ per unit of $x$ : metres per second, dollars per unit, metres per hour.
• Average rate on $[a, a + h]$ : $\frac{f(a+h) - f(a)}{h}$ , the slope of a secant. Instantaneous rate: its limit, the slope of the tangent.
• $f'(a) > 0$ : $f$ rising at $a$ , whatever the sign of $f(a)$ . $f'(a) = 0$ : horizontal tangent, often a peak or a valley, never the value $0$ .
• Sketching $f'$ from $f$ : plot the SLOPES of $f$ . A corner of $f$ becomes a gap in $f'$ , with open dots on both sides.

The slopes of

|x - 1|

are

-1

on the left of

1

and

1

on the right: the corner of

f

becomes a gap in the graph of

f'

, with an open dot on each side.

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 rewriting, by the form of f

Read a line as: for a function of the first form, the difference quotient is rewritten by the gesture of the second column, and the limit of the third follows. The red line is not a gesture problem: the two one-sided limits disagree, and no rewriting will make them agree.

Function and point	Gesture on the quotient	Result
$x^2 - 4x + 5$ at $3$	expand, cancel $h$	$2$
Example: $\frac{f(3+h) - f(3)}{h} = \frac{2h + h^2}{h} = 2 + h \to 2$
$x^3 - x$ at $2$	factor $(x - 2)$	$11$
Example: $\frac{x^3 - x - 6}{x - 2} = x^2 + 2x + 3 \to 4 + 4 + 3 = 11$
$\sqrt{2x - 1}$ at $5$	conjugate	$\frac{1}{3}$
Example: $\frac{\sqrt{9 + 2h} - 3}{h} = \frac{2}{\sqrt{9 + 2h} + 3} \to \frac{2}{6} = \frac{1}{3}$
$\frac{3}{x + 1}$ at $2$	common denominator	$-\frac{1}{3}$
Example: $\frac{\frac{3}{x+1} - 1}{x - 2} = \frac{2 - x}{(x + 1)(x - 2)} = -\frac{1}{x + 1} \to -\frac{1}{3}$
$\|x^2 - 4\|$ at $2$	one-sided quotients	no derivative does not exist
Example: From the right $\frac{4h + h^2}{h} \to 4$ , from the left $\frac{-4h - h^2}{h} \to -4$ . Same form, other result: $x\|x\|$ at $0$ : the quotient is $\|h\| \to 0$ from both sides, so $f'(0) = 0$ exists. What to do: Remove the absolute value by cases on each side of $a$ , compute both one-sided limits, and compare them.

In every green line the numerator of the rewritten quotient contains the factor $h$ (or $x - a$ ) exactly once: when it does not, a bracket was dropped in $f(a + h)$ .

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. Substituting h = 0 in the difference quotient

the whole question

What not to write

“ $f'(3) = \frac{f(3 + 0) - f(3)}{0} = \frac{0}{0}$ , so $f'(3)$ does not exist.”

What to write

“For $h \ne 0$ , $\frac{f(3+h) - f(3)}{h} = \frac{2h + h^2}{h} = 2 + h$ , so $f'(3) = \lim_{h\to 0}(2 + h) = 2$ .”

Why: $\frac{0}{0}$ is the shape of EVERY difference quotient at $h = 0$ ; it is the reason the derivative is a limit. The definition excludes $h = 0$ , and that is exactly what makes the cancellation of $h$ legitimate.

2. Applying f to only part of x + h

2 marks, and a wrong derivative for the rest of the question

What not to write

“ $f(x) = x^3 - 6x$ , so $f(x + h) = (x + h)^3 - 6x$ , and $f'(x) = 3x^2$ .”

What to write

“ $f(x + h) = (x + h)^3 - 6(x + h)$ , so $\frac{f(x+h) - f(x)}{h} = 3x^2 + 3xh + h^2 - 6 \to 3x^2 - 6$ .”

Why: $f(x + h)$ means: replace EVERY $x$ in the formula by $(x + h)$ , with brackets. The check is immediate: the terms without $h$ must cancel completely in $f(x + h) - f(x)$ .

3. Using the power rule when the question says by the definition

almost all the marks of the part, the number earning at most one

What not to write

“ $f(x) = \sqrt{2x - 1}$ , $f'(5) = \frac{2}{2\sqrt{9}} = \frac{1}{3}$ by the rules.”

What to write

“ $\frac{f(5+h) - f(5)}{h} = \frac{\sqrt{9 + 2h} - 3}{h} = \frac{2}{\sqrt{9 + 2h} + 3}$ after multiplying by the conjugate, so $f'(5) = \frac{1}{3}$ .”

Why: By the definition asks for the METHOD, the limit of a quotient; the rules of the next chapter serve only to check the answer. The same number reached without the limit answers another question.

4. Losing the sign in a quotient with a fraction

1 to 2 marks

What not to write

“ $\frac{\frac{3}{x+1} - 1}{x - 2} = \frac{x - 2}{(x+1)(x-2)} = \frac{1}{x + 1}$ , so $k'(2) = \frac{1}{3}$ .”

What to write

“ $\frac{3}{x+1} - 1 = \frac{3 - x - 1}{x + 1} = \frac{2 - x}{x + 1} = -\frac{x - 2}{x + 1}$ , so the quotient is $-\frac{1}{x + 1}$ and $k'(2) = -\frac{1}{3}$ .”

\frac{3}{x + 1}

decreases, so its tangent at

2

goes down with slope

-\frac{1}{3}

(green). The red line of slope

\frac{1}{3}

cannot touch a falling curve.

Why: $2 - x$ and $x - 2$ are opposites. The graph decides in two seconds: $\frac{3}{x+1}$ decreases for $x > -1$ , so its derivative there must be negative.

5. Calling a continuous function differentiable

the whole question

What not to write

“ $p(x) = |x^2 - 4|$ is continuous at $2$ , so it is differentiable at $2$ .”

What to write

“The quotient tends to $4$ from the right and to $-4$ from the left: the one-sided derivatives differ, so $p$ is not differentiable at $2$ (corner), although it is continuous.”

|x^2 - 4|

is unbroken at

2

, but it arrives with slope

-4

and leaves with slope

4

: a corner, so no derivative there.

Why: Differentiable implies continuous, never the reverse. Continuity says the graph has no break; differentiability says it has no break in its SLOPE either.

6. Taking a chord for a tangent

2 to 3 marks

What not to write

“The stone falls $s(t) = 4.9t^2$ m; at $t = 2$ it has fallen $19.6$ m in $2$ s, so its velocity at $t = 2$ is $9.8$ m/s.”

What to write

“ $\frac{s(2+h) - s(2)}{h} = 19.6 + 4.9h \to 19.6$ , so the velocity at $t = 2$ is $19.6$ m/s; $9.8$ m/s is the average velocity on $[0, 2]$ .”

t = 2

the chord from the origin has slope

9.8

m/s, the tangent has slope

19.6

m/s: on a curve bending upward, the chord is always less steep.

Why: $\frac{s(2)}{2}$ is the slope of the chord from the origin, an average over two seconds. The velocity at an instant is the slope of the tangent, a limit of averages over shorter and shorter intervals.

7. Rescaling the increment wrongly

the whole part

What not to write

“ $\lim_{h\to 0} \frac{f(3 + 2h) - f(3)}{h}$ has the shape of a derivative, so it equals $f'(3)$ .”

What to write

“With $k = 2h$ : $\frac{f(3 + 2h) - f(3)}{h} = 2 \cdot \frac{f(3 + k) - f(3)}{k} \to 2f'(3)$ .”

y = x^2 - 2x

the rise is taken over a run of

2h

; dividing it by

h

doubles the slope, so the limit is

2f'(3) = 8

, not

f'(3) = 4

Why: The denominator must be EXACTLY the increment added to $a$ . With $f'(3) = 4$ , the answer is $8$ , and $\lim_{h\to 0} \frac{f(3 - h) - f(3)}{h}$ is $-4$ .

8. Reading a rate as an amount

1 to 2 marks per sentence of interpretation

What not to write

“ $C'(100) = 4$ dollars per unit, so $100$ units cost $400$ dollars.”

What to write

“ $C'(100) = 4$ means that at a production of $100$ units the cost grows at about $4$ dollars per extra unit; the cost of $100$ units is $C(100) = 2300$ dollars.”

The cost of

100

units is a HEIGHT,

C(100) = 2300

dollars (green);

C'(100) = 4

dollars per unit is the SLOPE of the tangent (orange). Neither can be read from the other.

Why: A derivative has the units of $f$ PER unit of $x$ , and it describes one instant. To get an amount you need the function itself, never the rate multiplied by the variable.

Which method to choose

Which gesture, by the form of the difference quotient

Look at f before writing the quotient: its form picks the rewriting

If a polynomial → expand $f(a + h)$ fully, cancel the constant terms, factor out $h$

Example: $\frac{(2+h)^5 - 32}{h} = 80 + 80h + 40h^2 + 10h^3 + h^4 \to 80$
If a polynomial with the $x \to a$ form → $x = a$ is a root of the numerator: divide it by $(x - a)$

Example: $\frac{t^4 + t - 2}{t - 1} = t^3 + t^2 + t + 2 \to 5$
If a square root → multiply top and bottom by the conjugate, keep the bottom factored

Example: $\frac{\sqrt x - 3}{x - 9} = \frac{1}{\sqrt x + 3} \to \frac{1}{6}$
If a fraction → common denominator inside the numerator, then watch the sign of $a - x$

Example: $\frac{x}{2x+1}$ : numerator $h$ over $(2x + 2h + 1)(2x + 1)$ , so $f'(x) = \frac{1}{(2x+1)^2}$
If an absolute value, a piecewise function at a joint, or a root at an endpoint → compute the two one-sided quotients separately and compare them

Example: $\sqrt{4 - x}$ at $4$ : only $h < 0$ , quotient $-\frac{1}{\sqrt{-h}} \to -\infty$ , no derivative
If a limit already written, like $\lim \frac{\text{something} - \text{number}}{h}$ → recognize $f$ and $a$ : check that the number is $f(a)$ and the denominator is the increment

Example: $\lim_{x\to \pi/4} \frac{\tan x - 1}{x - \pi/4} = f'(\frac{\pi}{4})$ with $f = \tan$

In this chapter the differentiation rules are never a branch: they are the CHECK at the end. If no branch applies, go back to the definition of $f$ and write $f(a + h)$ with brackets.

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.

Computing f'(a) by the definition

When to use it: Any question that says use the definition, from first principles, or as a limit

1 Compute $f(a)$ on its own line, as a number.
2 Write $f(a + h)$ by replacing EVERY $x$ by $(a + h)$ , with brackets, and simplify it.
3 Write the quotient $\frac{f(a+h) - f(a)}{h}$ with its limit symbol, and name the form $\frac{0}{0}$ .
4 Rewrite for $h \ne 0$ (expand, conjugate, common denominator) until $h$ cancels, keeping $\lim$ on every line.
5 Let $h \to 0$ , state $f'(a)$ , and check it with the rules in one line if they are known.

Concluding sentence

“For $h \ne 0$ , $\frac{f(-1+h) - f(-1)}{h} = \frac{-7h + 2h^2}{h} = -7 + 2h$ , so $f'(-1) = \lim_{h\to 0}(-7 + 2h) = -7$ .”

The trap: Forgetting to say $h \ne 0$ when cancelling, or dropping $\lim$ in the middle lines: the computation is right and the write-up false.

Marking: Typically 1 mark for f(a) and f(a + h), 2 for the rewriting, 1 for the limit, and nothing for a number obtained by the rules alone.

Proving that f is not differentiable at a

When to use it: A corner, a joint between two formulas, an absolute value, a root that vanishes: any question that asks whether $f'(a)$ exists

1 Check continuity at $a$ first: if $f$ is not continuous there, conclude at once, since differentiable implies continuous.
2 Write the quotient for $h > 0$ with the formula valid on the right of $a$ , and compute its limit.
3 Write the quotient for $h < 0$ with the formula valid on the left, and compute its limit.
4 Compare: different finite values give a corner; infinite values give a vertical tangent or a cusp.

Concluding sentence

“ $\lim_{h\to 0^+} \frac{p(2+h) - p(2)}{h} = 4$ and $\lim_{h\to 0^-} \frac{p(2+h) - p(2)}{h} = -4$ . The one-sided derivatives are different, so $p'(2)$ does not exist: the graph has a corner at $(2, 0)$ .”

The trap: Differentiating each formula with the rules and comparing the results at $a$ : it works only when $f$ is continuous at $a$ , and it does not prove anything by itself in this chapter.

Marking: Typically 1 mark for continuity, 2 for each one-sided limit, 1 for the conclusion with the name of the defect.

Check before you hand in

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

The terms without h cancel

In $f(a + h) - f(a)$ , every term free of $h$ must disappear. If a constant or an $x^2$ survives, a bracket was dropped in $f(a + h)$ .

$(x + h)^3 - 6x - (x^3 - 6x) = 3x^2h + 3xh^2 + h^3$ : the missing $-6h$ reveals that $-6(x + h)$ was written $-6x$ .
The rules agree

Once the limit is done, check with the power rule in one line. It is not the method, it is the alarm.

By the definition $m'(2) = 11$ for $m(x) = x^3 - x$ ; the rule gives $3 \cdot 4 - 1 = 11$ .
The sign matches the graph

A rising graph has a positive derivative, a falling one a negative derivative. Look at f for two seconds.

$\frac{3}{x+1}$ falls for $x > -1$ , so $k'(2) = +\frac{1}{3}$ is impossible: the sign of $2 - x$ was lost.
The units are a rate

Write the units of f over the units of x after every derivative in a word problem. An answer in metres or in dollars alone is a value, not a derivative.

$s'(2) = 19.6$ m/s, $C'(100) = 4$ dollars per unit, $L'(4) \approx 0.35$ m/h.

The typical problem, taken apart

Two formulas, one joint: is the function differentiable there?

Let $f(x) = x^2 - 2x$ for $x \le 3$ and $f(x) = 4x - 9$ for $x > 3$ . Is $f$ differentiable at $3$ ? If so, give $f'(3)$ and the formula of $f'(x)$ for all $x$ .

No calculator, and no differentiation rule at the joint: every step by the definition, as on a MATH 140 midterm.

The parabola and the line meet at

(3, 3)

; the question is whether they arrive there with the same slope, which the picture alone cannot settle.

Step 1

Continuity: $f(3) = 9 - 6 = 3$ and $\lim_{x\to 3^+}(4x - 9) = 3$ , so $f$ is continuous at $3$ .

Why

If the pieces did not meet, the answer would be no at once, since differentiable implies continuous. Checking it first can save the whole computation.

Step 2

From the left, $h < 0$ : $\frac{f(3+h) - f(3)}{h} = \frac{(3+h)^2 - 2(3+h) - 3}{h} = \frac{4h + h^2}{h} = 4 + h \to 4$ .

Why

For $h < 0$ the point $3 + h$ is on the parabola, so $f(3 + h)$ uses the first formula, and $f(3)$ also, since $3$ belongs to the first piece.

Step 3

From the right, $h > 0$ : $\frac{f(3+h) - f(3)}{h} = \frac{4(3 + h) - 9 - 3}{h} = \frac{4h}{h} = 4$ .

Why

The trap is here: $f(3)$ is still $3$ , the value from the FIRST formula, even though $3 + h$ is on the line. The value at $a$ never changes side.

Step 4

Both one-sided limits exist, are finite and are equal to $4$ : $f$ is differentiable at $3$ and $f'(3) = 4$ .

Why

Naming the three conditions, exist, finite, equal, is where the method mark is. A joint is not automatically a corner.

Step 5

Away from $3$ , the definition gives $f'(x) = 2x - 2$ for $x < 3$ (quotient $2x - 2 + h$ ) and $f'(x) = 4$ for $x > 3$ . With $f'(3) = 4 = 2 \cdot 3 - 2$ : $f'(x) = 2x - 2$ for $x \le 3$ and $4$ for $x > 3$ .

Why

The formula of $f'$ is a piecewise function too, and the value at the joint must be placed in one of the pieces, here consistent with both.

The conclusion, written out

“ $f$ is continuous at $3$ , and $\lim_{h\to 0^-} \frac{f(3+h) - f(3)}{h} = \lim_{h\to 0^+} \frac{f(3+h) - f(3)}{h} = 4$ . Hence $f$ is differentiable at $3$ , $f'(3) = 4$ , and $f'(x) = 2x - 2$ for $x \le 3$ , $f'(x) = 4$ for $x > 3$ .”

The classic mistake on this problem: Using $f(3) = 4 \cdot 3 - 9$ in the right-hand quotient happens to work here, because the pieces meet; with a jump it would hide the discontinuity and give a finite slope for a function that has none.

Learn by heart

• $f'(a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h} = \lim_{x\to a} \frac{f(x) - f(a)}{x - a}$ : a $\frac{0}{0}$ form, rewritten before the limit.
• Tangent at $a$ : $y = f(a) + f'(a)(x - a)$ .
• Differentiable at $a$ : one-sided limits of the quotient EXIST, are FINITE and are EQUAL.
• Corner, cusp, vertical tangent, discontinuity: the four ways to have no derivative.
• Differentiable implies continuous; continuous does NOT imply differentiable ( $|x|$ at $0$ ).
• $f'(a)$ is a rate: units of $f$ per unit of $x$ . Secant: average rate. Tangent: instantaneous rate.
• By the definition means by the limit; the rules only check.

Frequently asked questions

How do I find the derivative using the limit definition?

Compute f of a, then f of a plus h by replacing every x with a plus h in brackets. Subtract, divide by h, and rewrite the quotient until the factor h cancels: expand a polynomial, multiply a square root by its conjugate, or put fractions over a common denominator. Only then let h go to zero, and write the limit symbol on every line until that last step.

Why is the difference quotient always 0 over 0?

Because at h equal to zero the numerator is f of a minus f of a, which is zero for every function, and the denominator is zero too. That form carries no information, which is exactly why the derivative is defined as a limit. The definition never sets h equal to zero: it simplifies the quotient for h different from zero, then lets h approach zero.

Can a function be continuous but not differentiable?

Yes. The absolute value of x is continuous at zero, but its slope is minus one on the left and plus one on the right, so it has a corner and no derivative there. The cube root of x is continuous at zero and has a vertical tangent, so no derivative either. The implication only goes one way: a differentiable function is always continuous.

What is the difference between average and instantaneous rate of change?

The average rate of change over an interval is the change in the function divided by the change in the variable, the slope of a secant. The instantaneous rate at a point is the limit of those averages as the interval shrinks to that point, the slope of the tangent: the derivative. For a falling object, the first is the average velocity over a few seconds, the second is the velocity at one instant.

What are the units of a derivative?

The units of the function divided by the units of the variable. If a distance in metres depends on a time in seconds, its derivative is in metres per second; if a cost in dollars depends on a number of units, its derivative is in dollars per unit. A derivative is a rate at one instant, so it is never an amount on its own.

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Practise it

Corrected exercises: The derivative as a limit, MATH 140 at McGill

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10 corrected exercises
100 points
150 minutes

Do the exercises

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Previous sheet Limits at infinity and asymptotes Next sheet Differentiation rules

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

Revision sheet: the derivative as a limit, and as a function (MATH 140)

The essentials

One definition, two forms, and a 0/0 form every time

When the derivative does not exist

What f'(a) means, in words and in units

The rules in table form

The rewriting, by the form of f

The mistakes that cost marks

1. Substituting h = 0 in the difference quotient

2. Applying f to only part of x + h

3. Using the power rule when the question says by the definition

4. Losing the sign in a quotient with a fraction

5. Calling a continuous function differentiable

6. Taking a chord for a tangent

7. Rescaling the increment wrongly

8. Reading a rate as an amount

Which method to choose

Which gesture, by the form of the difference quotient

How the answer is expected to be written

Computing f'(a) by the definition

Proving that f is not differentiable at a

Check before you hand in

The typical problem, taken apart

Two formulas, one joint: is the function differentiable there?

Learn by heart

Frequently asked questions

How do I find the derivative using the limit definition?

Why is the difference quotient always 0 over 0?

Can a function be continuous but not differentiable?

What is the difference between average and instantaneous rate of change?

What are the units of a derivative?

Corrected exercises: The derivative as a limit, MATH 140 at McGill

See also

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