Differential Calculus for Social Science 201-103 • CEGEP, Montreal

Revision sheet: the derivative and differentiation rules (201-103)

Q: How do I know which differentiation rule to apply?

Look at the last operation you would perform to compute the value on a calculator. If it is an addition, differentiate term by term. If it is a multiplication of two expressions containing the variable, it is the product rule. If it is a division, the quotient rule. If it is a power applied to a whole expression, it is the chain rule.

Q: Why do I have to multiply by the derivative of the inside?

Because a nested function passes its rate of change along in stages. If the inside changes three times as fast as the variable, the whole construction changes three times as fast too. The chain rule therefore multiplies the derivative of the outer layer by that of the inner layer, at every level of nesting.

Q: What are the units of a derivative?

The unit of the function divided by the unit of the variable. A cost in dollars differentiated with respect to a quantity gives dollars per unit. A rate in percent differentiated with respect to time gives percentage points per year. This mechanical calculation also works as a check: unreadable units signal a mistake in the rules.

Q: What does a negative second derivative mean?

That the rate of change is slowing down, not that the quantity is falling. If the first derivative is positive and the second negative, the quantity keeps increasing but more and more slowly. If both are negative, the quantity is falling and the fall is speeding up. Always read the two signs together.

Q: Is a continuous function always differentiable?

No, and it is a classic exam question. Differentiable implies continuous, but the converse is false. A tax schedule with brackets is continuous everywhere and yet not differentiable at the bracket boundaries, because the marginal rate changes abruptly there and creates a corner, so two different slopes at the same point.

Q: How do I estimate a derivative from a table of data?

Use the central difference: subtract the value before from the value after, then divide by twice the step. It is more accurate than the forward difference, because the errors on the two sides compensate. On a quadratic function, this estimate is even exact.

In Social Science, the derivative is almost never asked for on its own: you are asked what it is WORTH, what it is measured in, and what it says about the phenomenon. An answer that gives $f'(2)=9.44$ without writing “percentage points per year” loses half the question, even when the calculation is perfect.

This sheet gathers the nine mistakes that come back every exam period, the rules with a worked number under each, the two expected write-ups, and the units check, which on its own catches most wrong derivatives.

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, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

A derivative is not a formula, it is a SLOPE with units: until you have said what it is measured in and what it is the slope of, you have not answered the question asked.

This chapter is part of 201-103, Differential Calculus for Social Science

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

Average rate of change
Instantaneous rate of change
Derivative at a point
Product rule
Quotient rule
Chain rule
Equation of the tangent line
Second derivative
Relative rate of change
Differentiability and continuity
Central difference

The essentials

Two slopes, two questions

• AVERAGE rate between $a$ and $b$ : $\dfrac{f(b)-f(a)}{b-a}$ . It is the slope of the SECANT line, and it summarizes the whole interval.
• INSTANTANEOUS rate at $a$ : $f'(a)$ . It is the slope of the TANGENT line, and it describes only the point.
• Definition: $f'(a)=\lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}$ . The expression is not defined at $h=0$ , and it is the limit that fills the hole.
• For a QUADRATIC function ONLY, the average rate over an interval equals the instantaneous rate at the midpoint of that interval.
• In a table of data, the best estimate of $f'(a)$ is the CENTRAL difference $\dfrac{f(a+h)-f(a-h)}{2h}$ , not the forward difference.

The same curve, two slopes: on the left the line joins the points

a

and

b

and summarizes the interval, on the right it just grazes the single point

a

The statement always chooses its word: “on average over the period” asks for the secant, “at the moment when” asks for the tangent. Reading the word before calculating avoids half the mistakes.

The units, and what $f''$ really says

• $f'$ is measured in units of $f$ DIVIDED by units of $x$ : dollars per unit, percentage points per year, inhabitants per square kilometre.
• $f''$ is measured in units of $f$ divided TWICE by the unit of $x$ .
• $f'>0$ says that $f$ is rising. $f''<0$ says that $f$ is rising MORE SLOWLY than before, not that it is falling.
• “The increase is slowing down” is a statement about $f''$ . “The quantity is decreasing” is a statement about $f'$ . The two are independent.
• RELATIVE rate of change: $\dfrac{f'(t)}{f(t)}\times 100$ , in percent per unit of time. A constant absolute growth gives a DECREASING relative rate.

The slope of the tangent line at

t=2

is read on a triangle: one YEAR more,

9.44

POINTS more. The unit of

D'

is therefore the percentage point per year, the unit of

D

divided by the unit of

t

The units check is the fastest of the course: a derivative whose units make no sense signals a botched chain of rules, before you even reread the calculation.

Differentiable implies continuous, never the converse

• A function that is differentiable at $a$ is necessarily continuous there.
• The converse is false, and it is the one that gets tested: a function can be perfectly continuous and have no tangent line at all.
• Three ways to lose a derivative: the CORNER, where the two slopes differ; the JUMP, where the function is not even continuous any more; the VERTICAL TANGENT, where the slope becomes infinite.
• A marginal-rate schedule is continuous everywhere and not differentiable at the bracket boundaries: the marginal rate IS the slope, and changing the rate creates a corner.

The bracket schedule is the standard exam example, because it shows that a function without any jump can still fail to be differentiable.

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 basic rules, with a calculation under each

The last column gives the derivative. The red case marks a rule that does not exist and that exam papers nonetheless write every session.

Form of $f$	What you apply	$f'$
$5x^{3}$	the power rule	$15x^{2}$
Example: At $x=2$ : $f'(2)=15\times 4=60$ .
$\dfrac{1}{x^{2}}$	rewrite as $x^{-2}$ , then the power rule	$-\dfrac{2}{x^{3}}$
Example: At $x=2$ : $f'(2)=-\dfrac{2}{8}=-0.25$ .
$\sqrt{x}$	rewrite as $x^{1/2}$	$\dfrac{1}{2\sqrt{x}}$
Example: At $x=9$ : $f'(9)=\dfrac{1}{6}\approx 0.167$ .
$uv$	$u'v+uv'$	the sum of the two terms
Example: $x^{2}(3x+1)$ has derivative $9x^{2}+2x$ , which equals $40$ at $x=2$ .
$\dfrac{u}{v}$	$\dfrac{u'v-uv'}{v^{2}}$	the order is fixed
Example: $\dfrac{x}{x+1}$ has derivative $\dfrac{1}{(x+1)^{2}}$ , that is $\dfrac{1}{9}\approx 0.111$ at $x=2$ .
$(uv)'=u'v'$	nothing at all	does not exist rule that does not exist
Example: For $x^{2}\times x^{3}=x^{5}$ : the true derivative $5x^{4}$ equals $80$ at $x=2$ , whereas $2x\times 3x^{2}$ would give $48$ . What to do: Apply $u'v+uv'$ , or simplify the product before differentiating when that is possible.

The first three lines say the same thing: rewrite as a POWER before differentiating. This is the move that keeps you from inventing a rule for fractions and for roots.

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. Forgetting the derivative of the inner layer

2 marks, and it is the number one mistake of the course

What not to write

“ $f(x)=(3x+1)^{4}$ , so $f'(x)=4(3x+1)^{3}$ .”

What to write

“ $f'(x)=4(3x+1)^{3}\times 3=12(3x+1)^{3}$ : the chain rule multiplies by the derivative of the inside.”

x=1

, the true tangent line (green) has slope

4\times 4^{3}\times 3=768

. Without the factor

3

of the inner layer, you draw a line of slope

256

(red) that cuts the curve instead of touching it.

Why: Differentiating a nested function means walking through the layers from the outside in, multiplying at each step. Forgetting a factor amounts to answering a different question.

2. Reversing the order in the numerator of the quotient rule

the whole question, since the derivative obtained is the opposite of the true one

What not to write

“ $\left(\dfrac{u}{v}\right)'=\dfrac{uv'-u'v}{v^{2}}$ .”

What to write

“ $\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}}$ : the TOP is differentiated first.”

For

f(x)=\frac{x}{x+1}

, which RISES, the right formula gives

f'(2)=\frac{1}{9}>0

. The reversed numerator gives

-\frac{1}{9}

: a falling tangent line on a rising curve.

Why: The reversed sign gives a derivative that changes direction: an increasing function becomes decreasing, and the whole interpretation that follows is turned upside down.

3. Answering with the average rate when the instantaneous rate is asked

1 mark when the estimate is asked for, 3 when it is not

What not to write

“The rate of change at $t=2$ is $\dfrac{f(3)-f(1)}{2}$ .”

What to write

“The rate of change at $t=2$ is $f'(2)$ . The quantity $\dfrac{f(3)-f(1)}{2}$ is an ESTIMATE of this rate, the central difference, and it is exact only on a parabola.”

Why: The statement always distinguishes the two: “on average over the period” calls for the secant, “at the moment when” calls for the tangent. The word is in the question.

4. Reading $f''<0$ as a decline

2 interpretation marks, and it is the reasoning question of the chapter

What not to write

“The second derivative is negative, so the graduation rate is decreasing.”

What to write

“ $f'(2)=9.44>0$ and $f''(2)=-3.15<0$ : the graduation rate is still INCREASING, but more and more slowly.”

Both curves have a negative second derivative and do not tell the same story: the one on the left is still rising, it is only rising more and more slowly.

Why: $f'$ gives the direction, $f''$ says whether the movement speeds up or slows down. The two are independent, and an increase that slows down is still an increase.

5. Looking for the y-intercept to write a tangent line

1 mark, and two minutes lost per question

What not to write

“The slope is $9.44$ , so the tangent line is $y=9.44x+b$ , and I look for $b$ .”

What to write

“ $y=f(a)+f'(a)(x-a)$ , so $y=28.33+9.44(t-2)$ . The point-slope form gives the equation directly, with no need to find $b$ .”

Why: The point-slope form is made for this: it contains the point and the slope, the only two pieces of information you have. Expanding to find $b$ adds one more chance of making a sign error.

6. Giving a derivative without its units

half the question in Social Science, where the interpretation is graded

What not to write

“ $f'(2)=9.44$ .”

What to write

“ $f'(2)\approx 9.44$ percentage points per year: after two years, the graduation rate is rising by about $9.4$ points per additional year.”

Why: In Social Science, the number alone proves nothing. The units follow mechanically, the unit of $f$ divided by the unit of $x$ , and they also serve as a check on the calculation.

7. Believing that a continuous function is differentiable

2 marks, and it is the most frequently asked theory question of the chapter

What not to write

“The tax schedule is continuous, so it is differentiable everywhere.”

What to write

“The schedule is continuous everywhere, but not differentiable at the bracket boundaries: the marginal rate changes there, so the slope jumps, and there is a corner.”

Two of the three functions are continuous and none of the three is differentiable at the marked point: continuity never suffices to guarantee a tangent line.

Why: The implication works in one direction only. Differentiable implies continuous; continuous implies nothing, and the tax schedule is the official counterexample.

8. Confusing absolute growth and relative growth

2 interpretation marks

What not to write

“The population grows by $200$ inhabitants a year, so its growth rate is constant.”

What to write

“The ABSOLUTE increase is constant at $200$ inhabitants a year, but the RELATIVE rate $\dfrac{f'(t)}{f(t)}$ decreases, since the denominator grows.”

On the left, the same increase of

200

inhabitants every year: a line. On the right, what that increase weighs:

4\,\%

5{,}000

at the start,

2.2\,\%

9{,}000

twenty years later.

Why: The same gain weighs less and less on a growing base. This is exactly why the social sciences reason in percentages rather than in units.

9. Differentiating the top and the bottom separately, or the denominator alone

2 marks, and a derivative with the wrong sign

What not to write

“ $f(x)=\dfrac{1}{x^{2}}$ , so $f'(x)=\dfrac{1}{2x}$ .”

What to write

“Rewrite $f(x)=x^{-2}$ , then the power rule gives $f'(x)=-2x^{-3}=-\dfrac{2}{x^{3}}$ .”

The curve of

\frac{1}{x^{2}}

falls: its tangent line at

x=2

(green) has a NEGATIVE slope,

-0.25

. The wrong calculation

\frac{1}{2x}

gives

+0.25

, a line that rises: the sign alone was enough to reject it.

Why: Writing as a power before differentiating removes the need for a rule for fractions. The derivative must in any case be NEGATIVE, since the function is decreasing.

Which method to choose

Which rule, according to the LAST operation of the expression

Do not look at the topic, look at the operation you would perform last if you computed the value on a calculator. That operation names the rule.

If last, an addition or a subtraction → differentiate term by term, independently

Example: $3x^{2}-5x+7$ gives $6x-5$
If last, a multiplication of two expressions in $x$ → the product rule, $u'v+uv'$

Example: $x^{2}(3x+1)$ gives $9x^{2}+2x$

if the product expands easily, expanding first is often faster
If last, a division → the quotient rule, $\dfrac{u'v-uv'}{v^{2}}$ , in that order

Example: $\dfrac{85t}{t+4}$ gives $\dfrac{340}{(t+4)^{2}}$
If last, a power applied to a WHOLE expression → the chain rule: the power, then multiply by the derivative of the inside

Example: $(3x+1)^{4}$ gives $12(3x+1)^{3}$
If a constant multiplies the whole expression → take it out front, it is not differentiated

Example: $7\sqrt{x}$ gives $\dfrac{7}{2\sqrt{x}}$
If a fraction with no $x$ in the denominator, or a lone root → rewrite as a POWER before anything else

Example: $\dfrac{1}{x^{2}}$ becomes $x^{-2}$ , then $-2x^{-3}$

If several branches seem to apply, there are several layers: start with the last operation and work back down. No rule gets skipped, they nest.

Which quantity to compute, according to the wording of the statement

The vocabulary of Social Science is stable: each wording calls for one precise quantity, and only one.

If “on average between such and such a year” → average rate, so the secant $\dfrac{f(b)-f(a)}{b-a}$

Example: between 2020 and 2026, the average annual change
If “at the moment when”, “at the instant”, “in such and such a year” → instantaneous rate, so $f'(a)$ , with its units

Example: the speed of progress in 2026
If “by how many percent per year” → relative rate $\dfrac{f'(t)}{f(t)}\times 100$

Example: an increase of $9.44$ on a base of $28.33$ is $33.3$ percent
If “is the increase slowing down”, “is the phenomenon running out of steam” → sign of $f''$ , without changing the conclusion about $f'$

Example: $f'>0$ and $f''<0$ : the increase continues while slowing down
If “estimate the derivative from the table” → CENTRAL difference $\dfrac{f(a+h)-f(a-h)}{2h}$

Example: with $h=1$ : $\dfrac{f(3)-f(1)}{2}$
If “tangent line”, “linear approximation” → $y=f(a)+f'(a)(x-a)$ , in point-slope form

Example: $y=28.33+9.44(t-2)$

A statement that asks for “the change” with no further detail almost always asks for the instantaneous rate in CEGEP. If you are really in doubt, compute both and say which one answers the question.

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.

Writing the equation of a tangent line

When to use it: As soon as the statement mentions a tangent line or an approximating line, or asks you to estimate a nearby value from the derivative.

1 Compute $f(a)$ : this is the point of tangency, and it is computed with the function, not with the derivative.
2 Compute $f'(x)$ , then $f'(a)$ : this is the slope.
3 Write the point-slope form directly, $y=f(a)+f'(a)(x-a)$ , without looking for the y-intercept.
4 Expand ONLY if the statement asks for it.
5 Interpret the slope in the units of the context.

Concluding sentence

“The point of tangency is $\left(2, \dfrac{170}{6}\right)$ , that is about $(2, 28.33)$ , and the slope is $D'(2)=\dfrac{340}{36}\approx 9.44$ . The tangent line therefore has equation $y=28.33+9.44(t-2)$ .”

The trap: Computing the point of tangency with the derivative instead of the function. The error is silent: the equation obtained is a genuine line, just not the one asked for.

Marking: 1 mark for $f(a)$ , 2 marks for $f'(x)$ , 1 mark for $f'(a)$ , 1 mark for the equation written out.

Interpreting a derivative in context

When to use it: Every time a derivative calculation is followed by “interpret”, “what does it mean”, or “comment”. That is, almost always in 201-103.

1 Give the rounded numerical value, with the right number of decimals.
2 Give the UNITS, obtained by dividing the unit of $f$ by that of $x$ .
3 Write a complete sentence that names the quantity, the moment and the direction of change.
4 Add the sign of $f''$ when the statement speaks of speeding up or slowing down.
5 End with a caveat on validity when the model is a fit to data.

Concluding sentence

“ $D'(2)\approx 9.44$ percentage points per year: after two years of the program, the graduation rate is increasing by about $9.4$ points per additional year. Since $D''(2)\approx -3.15<0$ , this increase is slowing down.”

The trap: Writing “the derivative is $9.44$ , so it is increasing”. The sentence names neither the quantity, nor the moment, nor the unit: it earns nothing on the interpretation scheme.

Marking: 1 mark for the value, 1 mark for the units, 2 marks for the interpreted sentence.

Check before you hand in

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

The expected sign

Before rereading the calculation, ask yourself whether the function rises or falls at the point in question. The sign of the derivative you found must match.

$\dfrac{1}{x^{2}}$ is decreasing, so its derivative is negative. Getting $\dfrac{1}{2x}$ , which is positive at $x=2$ , signals the error without any recalculation.
The units

Write the units of $f$ divided by those of $x$ , and read the result aloud. If the sentence makes no sense, the chain of rules is wrong somewhere.

Dollars per unit produced reads well; dollars times units does not, and signals a multiplication where a division was needed.
The slope read on the graph

Plot the function on the calculator and look at the steepness at the point. A visibly gentle slope cannot equal $50$ .

A curve that barely gains one unit in ten signals a derivative of the order of $0.1$ , not $10$ .
The central difference as a numerical check

Compute $\dfrac{f(a+0.001)-f(a-0.001)}{0.002}$ on the calculator. The result must agree with the derivative you found to two or three decimals.

For $D(t)=\dfrac{85t}{t+4}$ at $t=2$ , the numerical calculation gives $9.4444$ , which confirms $\dfrac{340}{36}$ .
Consistency between $f'$ and $f''$

Reread the interpretation sentence and check that it does not say two contradictory things. An increase that slows down keeps $f'>0$ .

“The rate is decreasing faster and faster” requires $f'<0$ AND $f''<0$ . With $f'>0$ , the sentence is false whatever the calculation.

The typical problem, taken apart

The graduation rate: derivative, tangent line, units and relative rate

The graduation rate of a program, in percent, is modelled by $D(t)=\dfrac{85t}{t+4}$ , where $t$ is the number of years since the reform, with $t\ge 0$ .

Compute $D'(2)$ , interpret it with its units, write the tangent line at $t=2$ , then say whether the increase is speeding up or slowing down.

The graduation rate

D(t)=\frac{85t}{t+4}

over its first twenty years: this is the curve on which the point with

t=2

and its tangent line are placed.

Step 1

The last operation of $\dfrac{85t}{t+4}$ is a division: this is the quotient rule, with $u=85t$ and $v=t+4$ , so $u'=85$ and $v'=1$ .

Why

Naming $u$ , $v$ , $u'$ and $v'$ before writing the formula prevents the reversal of the numerator, which is the most frequent sign error of the chapter.

Step 2

$D'(t)=\dfrac{85(t+4)-85t\times 1}{(t+4)^{2}}=\dfrac{85t+340-85t}{(t+4)^{2}}=\dfrac{340}{(t+4)^{2}}$ .

Why

The numerator simplifies completely: this is the sign that the rule was applied in the right order. A numerator that keeps a $t$ signals an error.

Step 3

$D'(2)=\dfrac{340}{36}\approx 9.44$ . The derivative is positive for every $t$ , so the graduation rate keeps growing.

Why

The sign is read directly from the final form, without calculating: a square in the denominator and $340$ in the numerator never change sign.

Step 4

Units: $D$ is in percentage points, $t$ in years, so $D'$ is in percentage points PER YEAR.

Why

The units follow mechanically and serve as a check: they must read like a sentence of the context.

Step 5

Point of tangency: $D(2)=\dfrac{170}{6}\approx 28.33$ . Tangent line: $y=28.33+9.44(t-2)$ .

Why

The point comes from the FUNCTION, the slope from the derivative. Mixing the two produces a line that is correct but wrong for the question asked.

Step 6

$D''(t)=-\dfrac{680}{(t+4)^{3}}$ , so $D''(2)=-\dfrac{680}{216}\approx -3.15$ .

Why

The second derivative is computed by rewriting $D'(t)=340(t+4)^{-2}$ and applying the chain rule: the power comes down, and the sign becomes negative.

Step 7

$D'(2)>0$ and $D''(2)<0$ : the increase continues while SLOWING DOWN.

Why

This is the sentence the marking scheme looks for, and it combines the two signs. Concluding to a decline from $D''<0$ is the classic mistake.

Step 8

Relative rate: $\dfrac{D'(2)}{D(2)}=\dfrac{340/36}{170/6}=\dfrac{1}{3}$ , that is about $33.3$ percent per year.

Why

The relative rate relates the increase to its starting level. It is the reading expected in Social Science, because it can be compared from one program to another.

The conclusion, written out

“ $D'(2)=\dfrac{340}{36}\approx 9.44$ percentage points per year: two years after the reform, the graduation rate is progressing by about $9.4$ points per year, that is $33.3$ percent of its level. The tangent line at this point is $y=28.33+9.44(t-2)$ , and since $D''(2)\approx -3.15<0$ , the progress is slowing down.”

The classic mistake on this problem: Writing $D'(t)=\dfrac{85}{1}=85$ by differentiating the top and the bottom separately. The number obtained is plausible, it is in percent per year, and it is wrong by a factor of nine.

Learn by heart

• AVERAGE rate between $a$ and $b$ : slope of the secant line. INSTANTANEOUS rate at $a$ : $f'(a)$ , slope of the tangent line.
• Product: $u'v+uv'$ . Quotient: $\dfrac{u'v-uv'}{v^{2}}$ , the TOP is differentiated first.
• Chain: differentiate the outside, THEN multiply by the derivative of the inside. Never one without the other.
• Rewrite as a power before differentiating: $\dfrac{1}{x^{2}}=x^{-2}$ and $\sqrt{x}=x^{1/2}$ .
• Tangent line: $y=f(a)+f'(a)(x-a)$ . The point comes from $f$ , the slope from $f'$ .
• $f'$ is measured in units of $f$ per unit of $x$ . A derivative without units is an incomplete answer.
• $f''<0$ means “rising more slowly” or “falling faster”, never just “falling”.
• DIFFERENTIABLE implies CONTINUOUS. The converse is false: corner, jump, vertical tangent.

Frequently asked questions

How do I know which differentiation rule to apply?

Look at the last operation you would perform to compute the value on a calculator. If it is an addition, differentiate term by term. If it is a multiplication of two expressions containing the variable, it is the product rule. If it is a division, the quotient rule. If it is a power applied to a whole expression, it is the chain rule.

Why do I have to multiply by the derivative of the inside?

Because a nested function passes its rate of change along in stages. If the inside changes three times as fast as the variable, the whole construction changes three times as fast too. The chain rule therefore multiplies the derivative of the outer layer by that of the inner layer, at every level of nesting.

What are the units of a derivative?

The unit of the function divided by the unit of the variable. A cost in dollars differentiated with respect to a quantity gives dollars per unit. A rate in percent differentiated with respect to time gives percentage points per year. This mechanical calculation also works as a check: unreadable units signal a mistake in the rules.

What does a negative second derivative mean?

That the rate of change is slowing down, not that the quantity is falling. If the first derivative is positive and the second negative, the quantity keeps increasing but more and more slowly. If both are negative, the quantity is falling and the fall is speeding up. Always read the two signs together.

Is a continuous function always differentiable?

No, and it is a classic exam question. Differentiable implies continuous, but the converse is false. A tax schedule with brackets is continuous everywhere and yet not differentiable at the bracket boundaries, because the marginal rate changes abruptly there and creates a corner, so two different slopes at the same point.

How do I estimate a derivative from a table of data?

Use the central difference: subtract the value before from the value after, then divide by twice the step. It is more accurate than the forward difference, because the errors on the two sides compensate. On a quadratic function, this estimate is even exact.

Practise it

Corrected exercises: The derivative and differentiation rules, 201-103

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.

12 corrected exercises
120 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

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Revision sheet: the derivative and differentiation rules (201-103)

The essentials

Two slopes, two questions

The units, and what f′′f''f′′ really says

Differentiable implies continuous, never the converse

The rules in table form

The basic rules, with a calculation under each

The mistakes that cost marks

1. Forgetting the derivative of the inner layer

2. Reversing the order in the numerator of the quotient rule

3. Answering with the average rate when the instantaneous rate is asked

4. Reading f′′<0f''<0f′′<0 as a decline

5. Looking for the y-intercept to write a tangent line

6. Giving a derivative without its units

7. Believing that a continuous function is differentiable

8. Confusing absolute growth and relative growth

9. Differentiating the top and the bottom separately, or the denominator alone

Which method to choose

Which rule, according to the LAST operation of the expression

Which quantity to compute, according to the wording of the statement

How the answer is expected to be written

Writing the equation of a tangent line

Interpreting a derivative in context

Check before you hand in

The typical problem, taken apart

The graduation rate: derivative, tangent line, units and relative rate

Learn by heart

Frequently asked questions

How do I know which differentiation rule to apply?

Why do I have to multiply by the derivative of the inside?

What are the units of a derivative?

What does a negative second derivative mean?

Is a continuous function always differentiable?

How do I estimate a derivative from a table of data?

Corrected exercises: The derivative and differentiation rules, 201-103

See also

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The units, and what $f''$ really says

4. Reading $f''<0$ as a decline