MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: the derivative at a point and as a function (MATH 203)

This sheet is not a summary of sections 3.1 and 3.2 of Thomas' Calculus: you have the textbook. It answers one question, what makes students lose marks on the definition of the derivative in MATH 203 at Concordia University, and which precise gesture avoids each loss.

The idea of the chapter fits in one line, the limit of a difference quotient. The marks go elsewhere: in the algebra that rewrites that quotient before the limit. A scientific calculator is allowed on the exam, but it cannot take a limit; it can only confirm, with $h = 0.001$ , a number you found by hand.

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The thread of the chapter

$f'(a)$ is the limit of a quotient that is $\frac{0}{0}$ by construction, and the marks are lost in the ALGEBRA that rewrites it: the bracket around $f(a + h)$ , the complex fraction put over ONE common denominator, the sign of $a - x$ ; only once $h$ has cancelled does $h$ go to $0$ .

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

What is on this sheet

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

Topics covered

Derivative at a point
Difference quotient
Tangent line
Derivative as a function
One-sided derivatives
Differentiability
Differentiability and continuity
Corners and cusps
Vertical tangent
Instantaneous rate of change
Graph of the derivative

The essentials

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

• Slope of the secant through $P(a, f(a))$ and $Q(a + h, f(a + h))$ : $\frac{f(a + h) - f(a)}{h}$ . The tangent at $P$ is the limit of the secants as $h \to 0$ .
• $f'(a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h} = \lim_{x\to a} \frac{f(x) - f(a)}{x - a}$ : the same limit, with $x = a + h$ .
• At $h = 0$ the quotient is $\frac{f(a) - f(a)}{0} = \frac{0}{0}$ for every function: that form says nothing, so the quotient is REWRITTEN for $h \ne 0$ until $h$ cancels.
• As a function: $f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$ , with $x$ held fixed during the limit. Notations: $f'(x)$ , $\frac{dy}{dx}$ , $\frac{d}{dx}f(x)$ .
• Tangent at $a$ : $y = f(a) + f'(a)(x - a)$ .

y = \frac{1}{x}

, the secant from

P(2, \frac{1}{2})

Q(4, \frac{1}{4})

has slope

-\frac{1}{8}

; as

Q

slides to

P

the slopes

\frac{-1}{2(2 + h)}

reach the tangent's

-\frac{1}{4}

Keep $\lim$ on every line until the line where $h$ has disappeared. A line such as $\frac{-h}{h(2+h)} = \frac{-1}{2+h} = -\frac{1}{2}$ without $\lim$ is false as written.

When the derivative does not exist, and when it does

• Differentiable at $a$ : $\lim_{h\to 0^-}$ and $\lim_{h\to 0^+}$ of the quotient exist, are FINITE, and are EQUAL.
• Corner: two different finite slopes ( $|3 - x|$ at $3$ : $-1$ and $1$ ). Cusp: $-\infty$ on one side, $+\infty$ on the other.
• Vertical tangent: the quotient tends to $+\infty$ (or $-\infty$ ) on both sides ( $x^{1/5}$ at $0$ : $h^{-4/5} \to +\infty$ ).
• Discontinuity: no derivative, since differentiable implies continuous.
• NOT on the list: a change of formula, an absolute value, a fractional exponent. $x^{4/3}$ at $0$ : quotient $h^{1/3} \to 0$ , derivative $0$ .

Two fractional exponents, two verdicts: the curve with exponent

\frac{4}{3}

is flat at its dip (derivative

0

), the one with exponent

\frac{1}{5}

stands vertical (no derivative).

The algebra that decides for a power: $\frac{h^p}{h} = h^{p - 1}$ , which tends to $0$ if $p > 1$ and blows up if $p < 1$ .

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

• $f'(a)$ is a RATE at an instant, in units of $f$ per unit of the variable: degrees per minute, litres per minute, dollars per unit.
• Average rate on $[a, b]$ : $\frac{f(b) - f(a)}{b - a}$ , a secant. Instantaneous rate: $f'(a)$ , the tangent.
• $f'(a) > 0$ : $f$ rising at $a$ , whatever the sign of $f(a)$ . $f'(a) = 0$ : horizontal tangent $y = f(a)$ , not the $x$ -axis.
• Graph of $f'$ : plot the SLOPES of $f$ . A corner of $f$ is a gap in $f'$ , with an open dot on each side.

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. Concluding from 0/0

the whole question

What not to write

“ $f'(1) = \frac{f(1 + 0) - f(1)}{0} = \frac{0}{0}$ , so $f'(1)$ does not exist.”

What to write

“For $h \ne 0$ , $\frac{f(1 + h) - f(1)}{h} = \frac{-2h - h^2}{h} = -2 - h$ , so $f'(1) = \lim_{h\to 0}(-2 - h) = -2$ .” (for $f(x) = 5 - x^2$ )

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 $h = 0$ , which is what allows cancelling $h$ .

2. A minus sign that reaches only half of a square

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

What not to write

“ $f(x) = 5 - x^2$ , so $f(x + h) = 5 - x^2 + 2xh + h^2$ .”

What to write

“ $f(x + h) = 5 - (x + h)^2 = 5 - x^2 - 2xh - h^2$ , so the quotient is $-2x - h$ and $f'(x) = -2x$ .”

Why: $-(x + h)^2$ is minus the WHOLE square: expand the square inside its bracket first, then change every sign. The faulty line gives $f'(x) = +2x$ , a slope of the wrong sign for a parabola opening downward.

3. Subtracting fractions without a common denominator

the whole part

What not to write

“ $\frac{2}{x + h + 1} - \frac{2}{x + 1} = \frac{2 - 2}{h} = 0$ , so the derivative of $\frac{2}{x + 1}$ is $0$ .”

What to write

“ $\frac{2}{x + h + 1} - \frac{2}{x + 1} = \frac{2(x + 1) - 2(x + h + 1)}{(x + h + 1)(x + 1)} = \frac{-2h}{(x + h + 1)(x + 1)}$ , so the quotient tends to $-\frac{2}{(x + 1)^2}$ .”

Why: A complex fraction is simplified by ONE common denominator, then dividing by $h$ multiplies that denominator by $h$ . The check: after the common denominator, the numerator must be a multiple of $h$ .

4. Stopping after the common denominator when a root is left

2 to 3 marks

What not to write

“ $f(x) = \frac{2}{\sqrt{x + 1}}$ : $\frac{f(3 + h) - f(3)}{h} = \frac{2 - \sqrt{4 + h}}{h\sqrt{4 + h}}$ , which gives $\frac{0}{0}$ , so $f'(3)$ does not exist.”

What to write

“Multiply by the conjugate $2 + \sqrt{4 + h}$ : $\frac{4 - (4 + h)}{h\sqrt{4 + h}\,(2 + \sqrt{4 + h})} = \frac{-1}{\sqrt{4 + h}\,(2 + \sqrt{4 + h})} \to -\frac{1}{8}$ .”

Why: The common denominator removes the fraction, not the root: as long as the numerator is a difference of roots that tends to $0$ , the conjugate is still owed. Two gestures, in that order.

5. Differentiating the number f(a) instead of the function

the whole question

What not to write

“ $f(x) = x^2 + 4x$ , $f(-1) = -3$ , and a constant has derivative $0$ , so $f'(-1) = 0$ .”

What to write

“ $\frac{f(-1 + h) - f(-1)}{h} = \frac{2h + h^2}{h} = 2 + h$ , so $f'(-1) = 2$ .”

Why: $f'(-1)$ means: differentiate $f$ , THEN evaluate at $-1$ . Evaluating first turns the function into a number, and every number has derivative $0$ .

6. Calling a continuous function differentiable

the whole question

What not to write

“ $f(x) = |3 - x|$ has no break at $3$ , so it is differentiable at $3$ .”

What to write

“ $\frac{|3 - (3 + h)|}{h} = \frac{|h|}{h}$ is $1$ for $h > 0$ and $-1$ for $h < 0$ : the one-sided derivatives differ, so $f$ has a corner and $f'(3)$ does not exist.”

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

7. Reading the height of f as its slope

1 to 2 marks per interval

What not to write

“ $f(x) = 4 - x^2$ is positive on $(-2, 2)$ , so $f'(x) > 0$ on $(-2, 2)$ .”

What to write

“ $f'(x) = -2x$ (from the quotient $-2x - h$ ): $f' > 0$ on $(-2, 0)$ , where $f$ rises, and $f' < 0$ on $(0, 2)$ , where $f$ falls, although $f > 0$ there.”

(0, 2)

the parabola is above the axis while

f'(x) = -2x

is below it: positive height, negative slope.

Why: The sign of $f'$ is read as UP or DOWN on the graph of $f$ , never as above or below the axis.

8. Reading a rate as an amount

1 to 2 marks per sentence of interpretation

What not to write

“The oven $T(t) = 200 - \frac{720}{t + 4}$ has $T'(4) = 11.25$ , so after $4$ minutes it is $11.25$ degrees hotter than at the start.”

What to write

“At $t = 4$ min the temperature rises at $11.25$ degrees per minute; the change since the start is $T(4) - T(0) = 110 - 20 = 90$ degrees.”

Why: A derivative has the units of $f$ PER unit of the variable and describes one instant. An amount comes from $f$ itself, never from $f'$ alone.

Which method to choose

Which algebra gesture, by the form of f

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

If a polynomial, $h$ form → expand $f(a + h)$ with brackets, check that the terms without $h$ cancel, factor out $h$

Example: $4 - 3x - x^2$ at $-1$ : $f(-1 + h) = 6 - h - h^2$ , quotient $-1 - h \to -1$
If a polynomial, $x \to a$ form → $x = a$ is a root of the numerator: factor out $(x - a)$ , often by a difference of squares

Example: $\frac{x^4 - 81}{x - 3} = (x + 3)(x^2 + 9) \to 108$
If a square root → multiply top and bottom by the conjugate, keep the bottom factored

Example: $\frac{\sqrt{16 + 3h} - 4}{h} = \frac{3}{\sqrt{16 + 3h} + 4} \to \frac{3}{8}$
If a fraction → ONE common denominator, check the numerator is a multiple of $h$ , then divide by $h$

Example: $\frac{x + 1}{x - 1}$ at $3$ : numerator $4 + h - 2(2 + h) = -h$ , quotient $\frac{-1}{2 + h} \to -\frac{1}{2}$
If a root in a denominator → common denominator first, then the conjugate on what is left

Example: $\frac{4}{\sqrt{x + 12}}$ at $4$ : $\frac{-1}{\sqrt{16 + h}\,(4 + \sqrt{16 + h})} \to -\frac{1}{32}$
If an absolute value, a joint between two formulas, or an endpoint → two one-sided quotients, each with the formula valid on its side, then compare

Example: $|\sin x|$ at $0$ : $\frac{\sin h}{h} \to 1$ on the right, $-\frac{\sin h}{h} \to -1$ on the left

The differentiation rules of section 3.3 are never a branch in this chapter: they are the CHECK at the end. If no branch applies, go back to the formula and write $f(a + h)$ again, bracket by bracket.

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'(x) by the definition when f is a fraction

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

1 Write $f(x + h)$ by replacing EVERY $x$ by $(x + h)$ , with brackets.
2 Write $\lim_{h\to 0} \frac{f(x + h) - f(x)}{h}$ and name the form $\frac{0}{0}$ .
3 Put $f(x + h) - f(x)$ over ONE common denominator, expanding each product in full, with the minus sign on the whole second product.
4 Check that the new numerator is a multiple of $h$ ; cancel $h$ , saying $h \ne 0$ .
5 Let $h \to 0$ , state $f'(x)$ with its domain, and check it in one line with the calculator or with the rules if they are known.

Concluding sentence

“For $h \ne 0$ , $\frac{r(x + h) - r(x)}{h} = \frac{7}{(x + h + 3)(x + 3)}$ , so $r'(x) = \lim_{h\to 0} \frac{7}{(x + h + 3)(x + 3)} = \frac{7}{(x + 3)^2}$ for $x \ne -3$ .”

The trap: Subtracting the second product without brackets: $(2x + 2h - 1)(x + 3) - (2x - 1)(x + h + 3)$ must give exactly $7h$ ; any leftover $x$ or $hx$ term means a sign was lost.

Marking: Typically 1 mark for f(x + h), 2 for the common denominator and its numerator, 1 for the cancellation and the limit, and at most 1 for a correct answer obtained by the rules alone.

Deciding whether f is differentiable at a joint

When to use it: A piecewise function, an absolute value, or a question with parameters to choose so that $f'(a)$ exists

1 Compute $f(a)$ from the formula valid AT $a$ , and check continuity first: no continuity, no derivative.
2 For $h > 0$ , write the quotient with the formula valid on the right, and take its limit.
3 For $h < 0$ , write the quotient with the formula valid on the left, still with the same $f(a)$ , and take its limit.
4 Conclude: equal finite limits, differentiable; different finite limits, corner; infinite limits, vertical tangent or cusp.

Concluding sentence

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

The trap: Using, on each side, the value of $f(a)$ from that side's formula: it hides a jump and can produce a finite slope for a function that has none.

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 numerator is a multiple of h

After expanding or after the common denominator, every term of $f(a + h) - f(a)$ must contain $h$ . A surviving constant, $x^2$ or $x$ means a bracket was dropped.

$4 + h - 2(2 + h) = -h$ is right; $4 + h - 2 = 2 + h$ is not, and it was the missing bracket around $2 + h$ .
The calculator confirms with a small h

Compute $\frac{f(a + 0.001) - f(a)}{0.001}$ on the scientific calculator: it must be close to your $f'(a)$ . It checks the number, it never replaces the limit on the paper.

$m(x) = \frac{4}{\sqrt{x + 12}}$ : $\frac{m(4.001) - m(4)}{0.001} \approx -0.0312$ , against $-\frac{1}{32} = -0.03125$ . A result of $+0.03$ would betray a lost sign.
The sign matches the graph

A rising graph has a positive derivative, a falling one a negative derivative. Two seconds on the shape of f catch most sign errors.

$\frac{1}{x}$ decreases for $x > 0$ , so $\frac{d}{dx}\frac{1}{x}$ at $4$ cannot be $+\frac{1}{16}$ .
The units are a rate

In a word problem, write the units of f over the units of the variable after every derivative. An answer in degrees or in litres alone is a value, not a derivative.

$T'(4) = 11.25$ degrees per minute, $V'(10) = -200$ litres per minute.

The typical problem, taken apart

A fraction, by the definition, then its tangent

Let $f(x) = \frac{3}{2x - 1}$ . Find $f'(x)$ from the definition, give its domain, then write the tangent to the graph at $x = 2$ .

No differentiation rule: every step by the limit, as on a MATH 203 midterm; the calculator is used only to check.

The branch of

y = \frac{3}{2x - 1}

for

x > \frac{1}{2}

is decreasing, so every slope on it must come out negative, like the

-\frac{2}{3}

of the tangent at

(2, 1)

Step 1

$f(x + h) = \frac{3}{2(x + h) - 1} = \frac{3}{2x + 2h - 1}$ , and the quotient is $\frac{f(x + h) - f(x)}{h}$ , a $\frac{0}{0}$ form as $h \to 0$ .

Why

The bracket around $x + h$ is where the $2h$ comes from; writing $2x - 1 + h$ loses the factor $2$ and every later line with it.

Step 2

$\frac{3}{2x + 2h - 1} - \frac{3}{2x - 1} = \frac{3(2x - 1) - 3(2x + 2h - 1)}{(2x + 2h - 1)(2x - 1)} = \frac{-6h}{(2x + 2h - 1)(2x - 1)}$ .

Why

ONE common denominator, and the minus sign on the whole second product. The numerator $-6h$ is a multiple of $h$ : the check that the algebra is right.

Step 3

For $h \ne 0$ : $\frac{f(x + h) - f(x)}{h} = \frac{-6}{(2x + 2h - 1)(2x - 1)}$ , so $f'(x) = \lim_{h\to 0} \frac{-6}{(2x + 2h - 1)(2x - 1)} = \frac{-6}{(2x - 1)^2}$ , for $x \ne \frac{1}{2}$ .

Why

Dividing by $h$ multiplies the denominator by $h$ , which then cancels. The limit is written until $h$ has left the expression; the domain is part of the answer.

Step 4

At $x = 2$ : $f(2) = 1$ and $f'(2) = \frac{-6}{9} = -\frac{2}{3}$ . Tangent: $y = 1 - \frac{2}{3}(x - 2)$ , that is $y = -\frac{2}{3}x + \frac{7}{3}$ .

Why

The tangent needs a point AND a slope: $f(2)$ from $f$ , $f'(2)$ from $f'$ . Mixing them up, a line of slope $1$ , is a classic.

Step 5

Check: $\frac{f(2.001) - f(2)}{0.001} \approx -0.666$ on the calculator, and the sign is negative, like the decreasing branch.

Why

The calculator confirms the number in five seconds; it does not replace the limit, which is where the marks are.

The conclusion, written out

“For $h \ne 0$ , $\frac{f(x + h) - f(x)}{h} = \frac{-6}{(2x + 2h - 1)(2x - 1)}$ , hence $f'(x) = \frac{-6}{(2x - 1)^2}$ for $x \ne \frac{1}{2}$ , and the tangent at $x = 2$ is $y = -\frac{2}{3}x + \frac{7}{3}$ .”

The classic mistake on this problem: Writing the numerator $3(2x - 1) - 3(2x + 2h - 1)$ as $6x - 3 - 6x + 6h - 3$ : the minus sign reached one term, the numerator becomes $6h - 6$ , which is not a multiple of $h$ , and the check of step 2 catches it at once.

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 $a$ : $y = f(a) + f'(a)(x - a)$ .
• Fraction: ONE common denominator; the numerator must then be a multiple of $h$ .
• Root: conjugate. Root in a denominator: common denominator, THEN conjugate.
• Differentiable at $a$ : one-sided limits of the quotient EXIST, are FINITE and are EQUAL.
• Differentiable implies continuous; continuous does NOT imply differentiable.
• $f'(a)$ is a rate: units of $f$ per unit of the variable. $f'(a)$ differentiates $f$ first, evaluates at $a$ after.

Frequently asked questions

How do I find a derivative using the limit definition when the function is a fraction?

Write f of x plus h by replacing every x with x plus h in brackets. Subtract f of x and put the two fractions over one common denominator, expanding every product and putting the minus sign on the whole second product. The new numerator must be a multiple of h: cancel it, then let h go to zero. Keep the limit symbol on every line until h is gone.

What do I do when there is a square root in the denominator?

Two gestures in a row. First put the difference of the two fractions over one common denominator, which leaves a difference of roots on top. Then multiply top and bottom by the conjugate of that difference, so that the roots disappear from the numerator and a factor h appears. Cancel h, and only then take the limit.

Why does my difference quotient give 0 over 0?

Because it always does at h equal to zero: the numerator is f of a minus f of a, and the denominator is zero. That form carries no information and is not an answer. It is the reason the derivative is defined as a limit: you simplify the quotient for h different from zero until the factor h cancels, and only then let h approach zero.

Can a function be continuous but not differentiable?

Yes. The absolute value of x minus 3 is continuous at 3, but its slope is minus one on the left and plus one on the right, so it has a corner and no derivative there. A cusp or a vertical tangent gives the same verdict. The implication only goes one way: a function differentiable at a point is always continuous there.

What are the units of a derivative?

The units of the function divided by the units of the variable. A temperature in degrees Celsius that depends on a time in minutes has a derivative in degrees per minute; a volume in litres has a derivative in litres per minute. A derivative is a rate at one instant, so it is never an amount on its own.

Practise it

Corrected exercises: The derivative at a point and as a function, 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 Continuity Next sheet Differentiation rules

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math203-derivative-definition. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Get in touch for a first session. In MATH 203 the definition of the derivative is rarely misunderstood; it is the algebra inside the limit that costs the marks, and that is exactly what we practise.

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Revision sheet: the derivative at a point and as a function (MATH 203)

The essentials

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

When the derivative does not exist, and when it does

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

The mistakes that cost marks

1. Concluding from 0/0

2. A minus sign that reaches only half of a square

3. Subtracting fractions without a common denominator

4. Stopping after the common denominator when a root is left

5. Differentiating the number f(a) instead of the function

6. Calling a continuous function differentiable

7. Reading the height of f as its slope

8. Reading a rate as an amount

Which method to choose

Which algebra gesture, by the form of f

How the answer is expected to be written

Computing f'(x) by the definition when f is a fraction

Deciding whether f is differentiable at a joint

Check before you hand in

The typical problem, taken apart

A fraction, by the definition, then its tangent

Learn by heart

Frequently asked questions

How do I find a derivative using the limit definition when the function is a fraction?

What do I do when there is a square root in the denominator?

Why does my difference quotient give 0 over 0?

Can a function be continuous but not differentiable?

What are the units of a derivative?

Corrected exercises: The derivative at a point and as a function, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?