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

Revision sheet: implicit differentiation (201-103)

In Social Science, an implicit relation is almost never a geometric curve: it is an EQUILIBRIUM CONSTRAINT between two quantities that neither determines alone. An indifference curve, a production isoquant, a budget constraint, a possibilities frontier, a Phillips curve. The calculation is always the same, and it takes three steps: differentiate both sides treating $y$ as a function of $x$ , isolate $\dfrac{dy}{dx}$ , then READ the number you get in the vocabulary of the problem.

This sheet does not rehash the course: it lists what costs marks. The forgotten factor $\dfrac{dy}{dx}$ , which is the mechanical and fatal mistake of the chapter; the belief that the slope depends on $x$ alone; the vertical tangent confused with a zero derivative; the sign of the marginal rate, which is the OPPOSITE of the slope; and logarithmic differentiation left aside on a product of powers, which costs a page of calculations.

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The thread of the chapter

On an equilibrium constraint, $x$ and $y$ do not determine each other, they TRADE OFF. Every $y$ you differentiate leaves a factor $\dfrac{dy}{dx}$ behind, and the number you get is the rate of that trade, attached to a POINT and never to an abscissa alone.

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

Implicit differentiation
Chain rule
Product rule
Indifference curve
Isoquant
Marginal rate of substitution
Marginal rate of transformation
Logarithmic differentiation
Cobb-Douglas production function
Budget constraint
Vertical tangent
Consumer's optimum

The essentials

The move, in three steps

• An equilibrium constraint links $x$ and $y$ without either one determining the other. So you do not try to isolate $y$ : you look for the SLOPE at the bundle where you stand.
• Step 1: differentiate BOTH sides with respect to $x$ , treating $y$ as an unknown function of $x$ . The right-hand side is a constant, so its derivative is $0$ , and it is this zero that makes the isolation possible.
• Step 2: collect on one side all the terms that carry $\dfrac{dy}{dx}$ , factor $\dfrac{dy}{dx}$ out, divide.
• Step 3: evaluate at the given POINT, then read the number with its sign and the two units of the problem. A slope handed in without a sentence of interpretation never earns full marks.

The tangent at

(4, 9)

has a NEGATIVE slope,

-1.125

: the marginal rate of substitution is its opposite,

1.125

cultural outings given up for one more meal.

Writing out “ $y$ is a function of $x$ ” before the first line often earns half a mark for rigour, and above all it forces you to put the chain-rule factor in the right place.

The name of the number, according to the constraint

• On an indifference curve: $\text{MRS}=-\dfrac{dy}{dx}$ , the marginal rate of substitution, what the household is willing to give up of the second good for one unit of the first.
• On an isoquant: $\text{MRTS}=-\dfrac{dK}{dL}$ , the marginal rate of technical substitution, the machine-hours that one more worker replaces at constant output.
• On a production possibilities frontier: $\text{MRT}=-\dfrac{dy}{dx}$ , the marginal rate of transformation, that is, the opportunity cost.
• All three are POSITIVE because the slope is negative. A decreasing constraint is a trade-off: you cannot gain on both fronts.
• On a budget constraint $p_{x}x+p_{y}y=R$ , the slope is $-\dfrac{p_{x}}{p_{y}}$ . This is the trade the MARKET imposes, and it is the same at every point, unlike the marginal rate.

Indifference curve, isoquant, possibilities frontier: three DECREASING constraints, so three negative slopes and three positive rates (MRS, MRTS, MRT). Only the budget line keeps the same slope everywhere.

The word in the problem decides the name; the calculation is identical in all four cases. A student who writes “MRS” on a possibilities frontier shows that they learned a word, not a method.

Logarithmic differentiation, for products of powers

• As soon as the constraint is written $x^{a}y^{b}=c$ , take the logarithm BEFORE differentiating: $a\ln x+b\ln y=\ln c$ .
• Then differentiate with respect to $x$ , without forgetting the chain-rule factor on $\ln y$ : $\dfrac{a}{x}+\dfrac{b}{y}\,\dfrac{dy}{dx}=0$ .
• Hence $\dfrac{dy}{dx}=-\dfrac{a}{b}\cdot\dfrac{y}{x}$ , and the marginal rate equals $\dfrac{a}{b}\cdot\dfrac{y}{x}$ . Three lines, no fractional exponents.
• This is exactly the form of a Cobb-Douglas production function, the most common one in the whole economics course.

This is not another method: it is the same one, divided in advance by $x^{a}y^{b}$ , the quantity that gets in the way. Both routes give the same number, and one of them takes three lines.

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.

Each term, its derivative, and the factor it leaves

You differentiate with respect to $x$ , and $y$ is a function of $x$ . The right-hand column is what you write on your paper. The red cell is not an answer: it is the rule that does not exist.

Term	Chain-rule factor	Derivative with respect to $x$
$x^{3}$	no	$3x^{2}$
Example: At $x=2$ : $3\times 2^{2}=12$ .
$y^{3}$	yes	$3y^{2}\,\dfrac{dy}{dx}$
Example: At $y=2$ and $\dfrac{dy}{dx}=5$ : $3\times 4\times 5=60$ .
$xy$	yes, product	$y+x\,\dfrac{dy}{dx}$
Example: At $x=4$ , $y=2$ , $\dfrac{dy}{dx}=3$ : $2+4\times 3=14$ .
$x^{2}y$	yes, product	$2xy+x^{2}\,\dfrac{dy}{dx}$
Example: At $x=3$ , $y=5$ , $\dfrac{dy}{dx}=2$ : $30+18=48$ .
$y^{0.4}$	yes	$0.4\,y^{-0.6}\,\dfrac{dy}{dx}$
Example: At $y=32$ and $\dfrac{dy}{dx}=3$ : $0.4\times 0.125\times 3=0.15$ .
$\ln y$	yes	$\dfrac{1}{y}\,\dfrac{dy}{dx}$
Example: At $y=4$ and $\dfrac{dy}{dx}=8$ : $\dfrac{8}{4}=2$ .
$48$	no	$0$
Example: The right-hand side of $u^{2}\pi=48$ gives $0$ , and it is this zero that lets you isolate the slope.
$4y^{2}$	forgotten	$8y$ rule that does not exist
Example: On $x^{2}+4y^{2}=400$ , the equality $2x+8y=0$ would give $y=-3$ at $x=12$ , a negative production, whereas the point of the frontier is $(12, 8)$ . What to do: Write $\dfrac{d}{dx}\bigl(4y^{2}\bigr)=8y\,\dfrac{dy}{dx}$ , then isolate: $\dfrac{dy}{dx}=-\dfrac{x}{4y}$ , that is $-0.375$ at $(12, 8)$ .

Ask yourself a single question about each term: does $y$ appear in it? If so, a factor $\dfrac{dy}{dx}$ comes out of it, without exception.

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 $y^{2}$ as if $y$ were a number

the whole question

What not to write

“ $\dfrac{d}{dx}\bigl(4y^{2}\bigr)=8y$ , so $2x+8y=0$ and $y=-\dfrac{x}{4}$ .”

What to write

“ $y$ is a function of $x$ , so $\dfrac{d}{dx}\bigl(4y^{2}\bigr)=8y\,\dfrac{dy}{dx}$ ; the equation becomes $2x+8y\,\dfrac{dy}{dx}=0$ and $\dfrac{dy}{dx}=-\dfrac{x}{4y}$ .”

Without the factor

\frac{dy}{dx}

the calculation “finds”

y=-3

x=12

: off the frontier, a negative production. The true point is

(12, 8)

, where you look for a SLOPE.

Why: The wrong result remains a plausible expression, so nothing alerts you: at $x=12$ it announces $y=-3$ , a negative production, whereas the point of the frontier is $(12, 8)$ . Immediate check: if the calculation leaves no $\dfrac{dy}{dx}$ to isolate, the factor was forgotten.

2. Believing that the slope depends on $x$ alone

2 marks, and the interpretation that follows

What not to write

“I get $\dfrac{dw}{dh}=\dfrac{1}{4-0.2w}$ , but a $w$ is left in my answer: I must have made a mistake.”

What to write

“The slope is attached to a POINT, so it depends on $x$ AND $y$ : at $(30, 10)$ it equals $+0.50$ dollar per hour, and at $(30, 30)$ it equals $-0.50$ .”

Two points with the SAME abscissa

h=30

on one constraint: slope

+0.5

at the bottom,

-0.5

at the top. A slope belongs to a point, never to an abscissa.

Why: A constraint is not the graph of a function: the same abscissa can carry two points, so two slopes, often of opposite signs. Erasing the $y$ from the result means erasing the most useful information of the whole chapter.

3. Confusing a vertical tangent with a zero derivative

2 marks, and the economic conclusion with them

What not to write

“The tangent is vertical at this point, so $\dfrac{dy}{dx}=0$ .”

What to write

“The DENOMINATOR $4y$ vanishes at $(20, 0)$ , so $\dfrac{dy}{dx}$ does not exist and the tangent is vertical. It is horizontal at $(0, 10)$ , where it is the NUMERATOR that vanishes.”

The same frontier carries both cases: at

(0, 10)

the numerator vanishes and the slope is zero; at

(20, 0)

it is the denominator, and the derivative does not exist at all.

Why: The two cases are opposite and they do not say the same thing: horizontal tangent, the first unit costs almost nothing; vertical tangent, the economy is already fully specialized and no sacrifice buys the next unit.

4. Forgetting half of the product rule on a cross term

2 marks, and a quotient that stays wrong to the end of the exercise

What not to write

“ $\dfrac{d}{dx}\bigl(0.05xy\bigr)=0.05y$ , so $\dfrac{dy}{dx}=-\dfrac{2.8}{3}\approx -0.933$ .”

What to write

“ $\dfrac{d}{dx}\bigl(0.05xy\bigr)=0.05y+0.05x\,\dfrac{dy}{dx}$ , so $\dfrac{dy}{dx}=-\dfrac{2+0.05y}{3+0.05x}\approx -0.747$ at the bundle $(15, 16)$ .”

At the bundle

(15, 16)

, the correct tangent (green) has slope

-0.747

. By forgetting the piece

0.05x\,\frac{dy}{dx}

, you draw a steeper line (red,

-0.933

) that cuts the constraint instead of touching it.

Why: The error overstates the sacrifice by exactly 25 percent, and the result stays negative and plausible: nothing flags it. A term where $x$ and $y$ meet ALWAYS gives two pieces.

5. Announcing a negative marginal rate of substitution

1 mark of interpretation, on every question where the rate is asked

What not to write

“The slope is $-1.125$ , so the marginal rate of substitution is $-1.125$ .”

What to write

“The slope is $-1.125$ , and $\text{MRS}=-\dfrac{dy}{dx}=1.125$ : the rate is positive BECAUSE the slope is negative.”

Why: The minus sign in the definition is not decorative: it turns a trade-off into a quantity that can be compared with a price ratio. A rate announced as negative would say that the household gains on both goods at once.

6. Wanting to isolate $y$ before differentiating

the time of the exam, and often the whole question

What not to write

“I start from $3xy+y^{3}=28$ , I isolate $y$ , then I differentiate normally.”

What to write

“ $y$ cannot be isolated here, and the method never asks for it: I differentiate both sides, which gives $\dfrac{dy}{dx}=-\dfrac{y}{x+y^{2}}$ , that is $-0.1$ at the point $(9, 1)$ .”

To write

y

as a function of

x

, you would have to solve a cubic equation. Implicit differentiation gives the slope at the point

(9, 1)

without it:

-0.1

, the orange tangent.

Why: Isolating $y$ would here require solving a cubic equation for every $x$ . And even when $y$ can be isolated, the explicit route writes the slope as a function of $x$ alone, whereas the problem always gives you a POINT.

7. Differentiating a product of powers without taking the logarithm

2 marks, and half of the calculation errors of the chapter

What not to write

“ $0.3K^{-0.7}\,\dfrac{dK}{dL}\,L^{0.7}+0.7K^{0.3}L^{-0.3}=0$ ”, then three lines of exponents and a sign error.

What to write

“ $0.3\ln K+0.7\ln L=\ln 100$ , so $\dfrac{0.3}{K}\,\dfrac{dK}{dL}+\dfrac{0.7}{L}=0$ and $\dfrac{dK}{dL}=-\dfrac{7}{3}\cdot\dfrac{K}{L}$ .”

Why: The first write-up is correct, but isolating requires handling four fractional exponents of opposite signs. The logarithm turns the product into a sum and the exponents into coefficients: the calculation then takes three lines.

8. Handing in the number without reading it

2 marks out of 4, the interpretation always being graded separately

What not to write

“ $\dfrac{dK}{dL}=-2.33$ .” Full stop.

What to write

“ $\dfrac{dK}{dL}=-2.33$ : at the point $(100, 100)$ , one more worker replaces $2.33$ machine-hours at constant output. At $(150, 38.8)$ , the same worker replaces only $0.60$ .”

Two points of the SAME isoquant: the extra worker replaces

2.33

machine-hours at

(100, 100)

and only

0.60

(150, 38.8)

Why: In Social Science, the number is not the result: it is the trade-off. An answer without the two quantities named and without the direction of the sacrifice does not prove that you know what the derivative is talking about.

Which method to choose

Which route, according to the FORM of the constraint

Look at how the equation is written, not at the economic topic. Four forms cover the whole course.

If product of powers $x^{a}y^{b}=c$ → logarithm first, then $\dfrac{dy}{dx}=-\dfrac{a}{b}\cdot\dfrac{y}{x}$

Example: $K^{0.3}L^{0.7}=100$ gives $\dfrac{dK}{dL}=-\dfrac{7}{3}\cdot\dfrac{K}{L}=-2.333$ at $(100, 100)$

this is the form of a Cobb-Douglas function, so of almost every isoquant in the course
If cross term, $xy$ or $x^{2}y$ → product rule, TWO pieces, then collecting

Example: $x^{2}y=k$ gives $2xy+x^{2}\dfrac{dy}{dx}=0$ , so $\dfrac{dy}{dx}=-\dfrac{2y}{x}$

the forgotten piece is always the one that carries $\dfrac{dy}{dx}$
If sum of separate terms, $y$ appearing only once → differentiate directly, a single factor to isolate

Example: $2\sqrt{x}+y=20$ gives $\dfrac{dy}{dx}=-\dfrac{1}{\sqrt{x}}$ , that is $-0.5$ at $x=4$

here the slope does not contain $y$ , and that is normal: these are quasilinear preferences
If $y$ can be isolated cleanly → both routes work and give the SAME number

Example: $xy=36$ gives $-\dfrac{y}{x}$ or $-\dfrac{36}{x^{2}}$ , that is $-2.25$ at $(4, 9)$

the explicit route writes the slope in $x$ alone, the implicit one as a function of the point
If $y$ cannot be isolated, degree $3$ or more in $y$ → the implicit route is the only one open

Example: $3xy+y^{3}=28$ gives $-\dfrac{y}{x+y^{2}}=-0.1$ at $(9, 1)$

do not waste a minute looking for an explicit formula, in practice there is none

If no branch applies, differentiate both sides term by term anyway: the method never asks you to isolate $y$ , and it works on curves that are the graph of no function.

Which name does the number take, according to the WORD in the problem

The calculation is the same everywhere; it is the vocabulary of the problem that decides the name and the sentence of interpretation.

If “indifference curve”, “satisfaction”, “utility” → marginal rate of substitution, $\text{MRS}=-\dfrac{dy}{dx}$

Example: on $\sqrt{x}\,y=18$ , $\text{MRS}=\dfrac{y}{2x}=1.125$ at $(4, 9)$

expected sentence: what the household gives up of the second good for one unit of the first
If “isoquant”, “production”, capital and labour → marginal rate of technical substitution, $\text{MRTS}=-\dfrac{dK}{dL}$

Example: on $K^{0.3}L^{0.7}=100$ , $\text{MRTS}=2.333$ at $(100, 100)$

expected sentence: the machine-hours that one more worker replaces
If “possibilities frontier”, “opportunity cost” → marginal rate of transformation, $\text{MRT}=-\dfrac{dy}{dx}$

Example: on $x^{2}+4y^{2}=400$ , $\text{MRT}=\dfrac{x}{4y}=0.375$ at $(12, 8)$

it INCREASES along the frontier: this is the law of increasing opportunity costs
If prices and a budget are given → the slope of the budget is $-\dfrac{p_{x}}{p_{y}}$ , to be compared with the marginal rate

Example: at 3 dollars and 1 dollar per unit, the slope of the budget is $-3$ at every point

the optimum is the equality $\text{MRS}=\dfrac{p_{x}}{p_{y}}$ , and nothing else
If unemployment and inflation → sacrifice ratio, $\left|\dfrac{du}{d\pi}\right|$

Example: on $u^{2}\pi=48$ , the ratio is $\dfrac{u^{3}}{96}$ , that is $0.667$ at $u=4$ and $2.25$ at $u=6$

a ratio announced without the unemployment rate at which it was measured means nothing

In all five cases the number is announced as POSITIVE, because it is the opposite of a negative slope. And in all five cases it is attached to the point where it was computed: it cannot be extrapolated, exactly as a marginal cost is not multiplied by a hundred units.

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 and interpreting a marginal rate on a constraint

When to use it: The problem gives an equation linking two quantities and a point, and asks for “the rate” or “how much must be given up”.

1 Check that the point satisfies the equation, and write it out: $12^{2}+4\times 8^{2}=144+256=400$ .
2 State the convention: “ $y$ is a function of $x$ , I differentiate both sides with respect to $x$ .”
3 Differentiate term by term, putting a factor $\dfrac{dy}{dx}$ at every $y$ and two pieces at every cross term. The constant side gives $0$ : $2x+8y\,\dfrac{dy}{dx}=0$ .
4 Collect the terms in $\dfrac{dy}{dx}$ , factor, divide: $\dfrac{dy}{dx}=-\dfrac{x}{4y}$ .
5 Evaluate at the point, with the sign: $\dfrac{dy}{dx}=-\dfrac{12}{32}=-0.375$ .
6 Name the rate, give it as a positive value, and interpret it with BOTH units of the problem.

Concluding sentence

“At the point $(12, 8)$ , $\dfrac{dy}{dx}=-0.375$ , so the marginal rate of transformation is $0.375$ : to produce one more thousand tonnes of agricultural goods, this economy must give up 375 units of industrial goods.”

The trap: Skipping the check of the point. A slope computed at a point that is not on the constraint is worth nothing, and the grader sees it from the first line.

Marking: 1 mark for the differentiation with the chain-rule factor, 1 mark for the isolation, 1 mark for the signed value, 1 mark for the interpretation with units. The interpretation is always graded separately.

Finding the consumer's optimal bundle

When to use it: The problem gives a utility function, two prices and a budget, and asks for the chosen bundle.

1 Set the indifference curve $U(x, y)=k$ and compute its slope by implicit differentiation.
2 Deduce the marginal rate of substitution, the opposite of this slope.
3 Write the budget constraint $p_{x}x+p_{y}y=R$ and its slope, $-\dfrac{p_{x}}{p_{y}}$ .
4 Set the tangency condition $\text{MRS}=\dfrac{p_{x}}{p_{y}}$ and derive a relation between $x$ and $y$ .
5 Substitute this relation into the budget constraint, obtain the bundle, then compute the utility reached.

Concluding sentence

“At the optimum, $\text{MRS}=\dfrac{p_{x}}{p_{y}}$ , that is $\dfrac{3}{2}\cdot\dfrac{y}{x}=3$ , so $y=2x$ ; substituted into $3x+y=120$ , this gives the bundle $(24, 48)$ , which exactly exhausts the budget of 120 dollars.”

The trap: Handing in the relation $y=2x$ as the answer. It gives a DIRECTION, not a bundle: it is the substitution into the budget constraint that fixes the point.

Marking: 2 marks for the implicit slope, 1 mark for the slope of the budget, 1 mark for the tangency condition, 2 marks for the substitution and the bundle.

Check before you hand in

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

Is the point really on the constraint?

Before any differentiation, replace $x$ and $y$ by the values of the point in the starting equation. Ten seconds, and everything else depends on it.

On $x^{2}+4y^{2}=400$ , the point $(12, 8)$ gives $144+256=400$ : it is on it. A point that gave $398$ is not on the frontier, and the whole question will be wrong.
Is there still a $\dfrac{dy}{dx}$ to isolate?

Count the $\dfrac{dy}{dx}$ you put in after differentiating. If there is none, the chain-rule factor was forgotten somewhere and there is nothing left to isolate.

$2x+8y=0$ contains no slope unknown: it is the symptom of the omission, not an equation to solve.
Is the marginal rate positive?

On a decreasing constraint, the slope is negative and the marginal rate is its opposite, so POSITIVE. A rate announced as negative is a sign error, never a result.

A slope of $-1.125$ gives a rate of $1.125$ . A $-1.125$ handed in as the rate would say that the household gains on both goods at once.
Does the slope still contain $y$ ?

On a product of powers or a cross term, the slope always contains $y$ . A result in $x$ alone signals a lost factor, except in the quasilinear case, where $y$ appears only once and to the first degree.

$-\dfrac{a}{b}\cdot\dfrac{y}{x}$ keeps $y$ ; on the other hand $2\sqrt{x}+y=20$ gives $-\dfrac{1}{\sqrt{x}}$ , that is $-0.5$ at $x=4$ , without $y$ , and that is normal.
Does a neighbouring point confirm the slope?

Take a point near the constraint and compare the actual change with the one the slope predicts, $\Delta y\approx\dfrac{dy}{dx}\times\Delta x$ . It is the only possible check when $y$ cannot be isolated.

On $3xy+y^{3}=28$ , the slope at $(9, 1)$ is $-0.1$ , so $\Delta y\approx -0.0100$ for $\Delta x=0.1$ . The exact ordinate at $x=9.1$ is $0.9901$ , that is $\Delta y=-0.0099$ : confirmed.

The typical problem, taken apart

A household's optimal bundle, with a Cobb-Douglas utility

A household devotes 120 dollars a month to two expenses: $x$ , the number of delivered meals, at 3 dollars per unit, and $y$ , the number of public transit trips, at 1 dollar per unit. Its satisfaction is measured by $U=x^{0.6}y^{0.4}$ .

Find the bundle it chooses, then check the result. The figure shows the budget line and two levels of satisfaction: one crosses the budget, the other is out of reach.

The curve

U=28

CROSSES the budget line, so a better bundle exists; the curve

U=40

touches it nowhere, so it is out of reach. The optimum lies between the two.

Step 1

The household is indifferent along the curve $x^{0.6}y^{0.4}=k$ . Since $y$ is an unknown function of $x$ on it, we do not try to isolate it.

Why

Stating from the start that the curve is a constraint avoids the ten minutes lost isolating $y$ , and it is the sentence that earns the method mark.

Step 2

Take the logarithm of both sides: $0.6\ln x+0.4\ln y=\ln k$ .

Why

The constraint is a product of powers, the exact form where the logarithm turns the calculation into one line. Without it, four fractional exponents to isolate.

Step 3

Differentiate with respect to $x$ , with the chain-rule factor on $\ln y$ : $\dfrac{0.6}{x}+\dfrac{0.4}{y}\,\dfrac{dy}{dx}=0$ .

Why

This is where the chapter is won or lost: without this factor, the equality has no unknown left and the calculation stops on an impossibility.

Step 4

Isolate: $\dfrac{dy}{dx}=-\dfrac{0.6}{0.4}\cdot\dfrac{y}{x}=-\dfrac{3}{2}\cdot\dfrac{y}{x}$ , so $\text{MRS}=\dfrac{3}{2}\cdot\dfrac{y}{x}$ .

Why

The rate is the OPPOSITE of the slope, so positive. It depends on the ratio $\dfrac{y}{x}$ , that is, on the bundle, and not on the abscissa alone.

Step 5

The budget constraint is $3x+y=120$ , so $y=120-3x$ : the line has slope $-3$ , the opposite of the price ratio $\dfrac{p_{x}}{p_{y}}=\dfrac{3}{1}=3$ .

Why

This slope is the trade the market imposes, and it is the same at every point: it is the only object of the problem that does not depend on the bundle.

Step 6

At the optimum the two slopes coincide: $\dfrac{3}{2}\cdot\dfrac{y}{x}=3$ , so $y=2x$ .

Why

As long as the two trades differ, there is an exchange that improves the household without leaving the budget. The equality is therefore the condition, and it is written before any numerical calculation.

Step 7

Substitute into the budget: $3x+2x=120$ , so $x=24$ and $y=2\times 24=48$ . The utility reached is $24^{0.6}\times 48^{0.4}\approx 31.67$ .

Why

The substitution into the constraint is what turns a condition into a bundle. The condition alone never earns full marks.

Step 8

Check: $3\times 24+48=120$ , so the bundle is on the budget, and $\text{MRS}=\dfrac{3}{2}\times\dfrac{48}{24}=3$ , equal to the price ratio. The neighbouring bundle $(25, 45)$ also costs 120 dollars and gives only $31.63$ .

Why

Two independent checks in ten seconds: the budget, and a neighbour that must do WORSE. A better neighbour would prove that the condition was set up wrongly.

The conclusion, written out

The household chooses 24 delivered meals and 48 trips, which exhausts its 120 dollars and brings its satisfaction to $U\approx 31.67$ . At this bundle, it accepts giving up 3 trips for one more meal, exactly what the market asks of it, since $\text{MRS}=\dfrac{p_{x}}{p_{y}}=3$ .

The classic mistake on this problem: Setting the equality $\text{MRS}=\dfrac{p_{x}}{p_{y}}$ , obtaining $y=2x$ and handing in that relation as the answer. The tangency condition gives a direction, not a bundle: only the substitution into the budget constraint fixes the point.

Learn by heart

• Every $y$ you differentiate leaves a factor: $\dfrac{d}{dx}\bigl(y^{2}\bigr)=2y\,\dfrac{dy}{dx}$ , NEVER $2y$ .
• A term where $x$ and $y$ meet gives two pieces: $\dfrac{d}{dx}(xy)=y+x\,\dfrac{dy}{dx}$ .
• The right-hand side of a constraint is a CONSTANT, so its derivative is $0$ , and it is this zero that lets you isolate.
• For $x^{a}y^{b}=c$ : $\dfrac{dy}{dx}=-\dfrac{a}{b}\cdot\dfrac{y}{x}$ , so the marginal rate equals $\dfrac{a}{b}\cdot\dfrac{y}{x}$ .
• $\text{MRS}=-\dfrac{dy}{dx}$ : the rate is POSITIVE because the slope is NEGATIVE.
• Horizontal tangent: the NUMERATOR vanishes. Vertical tangent: the DENOMINATOR vanishes and the derivative does not exist.
• Consumer's optimum: $\text{MRS}=\dfrac{p_{x}}{p_{y}}$ , the trade accepted equals the trade imposed.
• An implicit slope belongs to a POINT: two points with the same abscissa can carry opposite slopes.

Frequently asked questions

Why do you have to add dy over dx when you differentiate y?

Because y is not an inert letter: on a constraint, y is an unknown function of x. The chain rule therefore requires you to differentiate the expression, then multiply by the derivative of the inside, which is precisely dy over dx. The derivative of y squared with respect to x equals two y times dy over dx, never just two y.

How do you compute a marginal rate of substitution?

Differentiate both sides of the indifference curve with respect to x, treating y as a function of x, isolate dy over dx, then evaluate it at the given bundle. The marginal rate of substitution is the opposite of that slope. On a product of powers x to the a times y to the b, it equals a over b multiplied by y over x.

Why is the marginal rate of substitution positive?

Because it is defined as the opposite of the slope, and the slope of an indifference curve is negative. That negative sign says you cannot increase both quantities and stay equally satisfied: you must give up some of one to get more of the other. The rate measures this sacrifice, and a sacrifice is stated as a positive number.

When should you take the logarithm before differentiating?

As soon as the constraint is a product of powers, such as a Cobb-Douglas production function. Write a times the logarithm of x plus b times the logarithm of y equals the logarithm of the constant, then differentiate: the product becomes a sum and the exponents become coefficients. Three lines instead of a page of fractional exponents.

What does a vertical tangent mean on an economics curve?

It appears where the denominator of the derivative vanishes, so where the derivative does not exist. It is not a zero derivative, which would give a horizontal tangent instead. In economics, a vertical tangent marks a ceiling: no change in the other quantity buys one more unit.

Practise it

Corrected exercises: Implicit differentiation, 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.

11 corrected exercises
110 points
150 minutes

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Revision sheet: implicit differentiation (201-103)

The essentials

The move, in three steps

The name of the number, according to the constraint

Logarithmic differentiation, for products of powers

The rules in table form

Each term, its derivative, and the factor it leaves

The mistakes that cost marks

1. Differentiating y2y^{2}y2 as if yyy were a number

2. Believing that the slope depends on xxx alone

3. Confusing a vertical tangent with a zero derivative

4. Forgetting half of the product rule on a cross term

5. Announcing a negative marginal rate of substitution

6. Wanting to isolate yyy before differentiating

7. Differentiating a product of powers without taking the logarithm

8. Handing in the number without reading it

Which method to choose

Which route, according to the FORM of the constraint

Which name does the number take, according to the WORD in the problem

How the answer is expected to be written

Computing and interpreting a marginal rate on a constraint

Finding the consumer's optimal bundle

Check before you hand in

The typical problem, taken apart

A household's optimal bundle, with a Cobb-Douglas utility

Learn by heart

Frequently asked questions

Why do you have to add dy over dx when you differentiate y?

How do you compute a marginal rate of substitution?

Why is the marginal rate of substitution positive?

When should you take the logarithm before differentiating?

What does a vertical tangent mean on an economics curve?

Corrected exercises: Implicit differentiation, 201-103

See also

Looking for a 201-103 tutor in Montreal?

1. Differentiating $y^{2}$ as if $y$ were a number

2. Believing that the slope depends on $x$ alone

6. Wanting to isolate $y$ before differentiating