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EECM 3714

Lecture 4: Unit 4
Differentiation

Renshaw, Ch
...
4-10,13
04 March 2022

OUTLINE
Renshaw, Ch
...
4-10 & 13
1
...
Differentiation rules
3
...
Examples
5
...

• Look at fig
...
1a and 6
...
When we move from P to Q, we measure the slope of 𝑦 = 𝑎𝑥 + 𝑏 as the
change in y, Dy, divided by the change in x, Dx
...


The diff
...
measure the slope, or rate of change of y as x
changes, between P and Q
• In fig
...
1a the diff
...
is positive because Dy is positive; in
fig
...
1b it is negative because Dy is negative
...
6
...
3

Δ𝑦
Δ𝑥

is the same in both, but the curves are very different
...
quot
...

• Another problem: the diff
...
also varies with distance from P to Q
...
4 to address this problem, we can use the slope of the tangent to the

curve at P as the measure of slope of curve at that point
...
5)
•

Δ𝑦
Δ𝑥

then approaches a limiting value, which

measures slope of tangent at P
...

∆𝑦
∆𝑥→0 ∆𝑥

• Derivative is lim

𝑑𝑦

= 𝑑𝑥

• So if 𝑦 = 𝑓(𝑥) then the slope of the function is
∆𝑦
∆𝑥→0 ∆𝑥

𝑓 ′ 𝑥 = lim

𝑑𝑦

= 𝑑𝑥

RULES OF DIFFERENTIATION
• “Differentiation” means finding derivative of a function
...

• For any function 𝑦 = f(𝑥), we write its derivative as:
either

d𝑦
d𝑥

or f ′ 𝑥

• The notation f′(𝑥) is obviously more compact
...
Power rule

y=x

n

2
...
Additive constant

y = f( x ) + B

4
...
Power rule: if 𝑦 = 𝑥 3 , we have 𝑛 = 3, so

d𝑦
d𝑥

= 3𝑥 3−1 = 3𝑥 2

2a
...
Multiplicative constant:if y = Ax, then

𝑑𝑦
𝑑𝑥

=𝐴

3
...
Sum or difference: 𝑦 = 𝑥 3 + 𝑥 2 ,

d𝑦
d𝑥

d𝑦
d𝑥

= 3𝑥 2

= 3𝑥 2 + 2𝑥

RULES OF DIFFERENTIATION II
5
...
Product

7
...
Inverse function

y = f(u ) where u = g( x )

y = uv

where u and v are
functions of x

u
y=
v

where u and v are
functions of x

y = f( x )

dy dy du
=
dx du dx
dy
dv
du
=u
+v
dx
dx
dx

dy
=
dx
dy
=
dx

v

du
dx

dv

− u dx
v2

1
dx
dy

EXAMPLES FOR RULES 5 − 8:
5
...
Product: given 𝑦 = (𝑥 2 + 1)(𝑥 3 + 𝑥 2 )
Create 2 new variables:𝑢 = 𝑥 2 + 1 and 𝑣 = 𝑥 3 + 𝑥 2
...
Quotient: given 𝑦 =

𝑥 2 +1
𝑥 3 +𝑥 2

Create 2 new variables: 𝑢 = 𝑥 2 + 1 and 𝑣 = 𝑥 3 + 𝑥 2
dv
d𝑢
2
So:
= 3𝑥 + 2𝑥 and
= 2x
d𝑥

Then:

d𝑥

d𝑦
d𝑥

d𝑢

=

d𝑣

𝑣 d𝑥 −𝑢d𝑥

𝑣2


...
Inverse function:

given 𝑦 = 𝑥 2 ,

d𝑦
d𝑥

= 2𝑥, so

d𝑥
d𝑦

=

1
d𝑦
d𝑥

=

1
2𝑥

Do examples 6
...
195-205 and all Progress exercise
...
399-401
• For logarithmic functions:
• If 𝑦 =

• If 𝑦 =
• Example:

𝑑𝑦
𝑑𝑥

1
ln 𝑥, then =
𝑥
𝑑𝑦
𝑓′(𝑥)
ln 𝑓(𝑥), then =
𝑑𝑥
𝑓(𝑥)

𝑦 = ln( 𝑥 2 + 3𝑥) ;

d𝑦
d𝑥

let 𝑢 = 𝑥 2 + 3𝑥 So:
𝑦 = ln( 𝑢)

=

2𝑥+3
𝑥 2 +3𝑥

d𝑦
d𝑥

• For exponential functions:
𝑥

• If 𝑦 = 𝑒 , then

𝑑𝑦
𝑑𝑥

d𝑢
d𝑥

𝑑𝑦
𝑑𝑥

• Example 1:

2 +3𝑥
𝑥
10𝑒

= 𝑓′(𝑥)𝑒 𝑓(𝑥)

• Example 2: 𝑦 = 𝑎𝑒 𝑟𝑥 ;

d𝑦
d𝑥

+ 50;

d𝑦
d𝑥

=

2 +3𝑥
𝑥
10𝑒
(2𝑥

= 𝑎𝑒 𝑟𝑥 (𝑟) = 𝑟𝑎𝑒 𝑟𝑥

1

=𝑢
2𝑥+3

= 𝑥 2 +3𝑥

= 𝑒𝑥

• If 𝑦 = 𝑒 𝑓(𝑥) , then
𝑦=

d𝑦

= d𝑢 𝑥

d𝑦
d𝑢

d𝑢
= 2𝑥 + 3
d𝑥

+ 3)

EXAMPLE 1
• Suppose that 𝑦 = 4𝑥 2 − 3 2𝑥 5 + 𝑥
• Find

𝑑𝑦
𝑑𝑥

• Solution

• Let 𝑢 = 4𝑥 2 − 3 and 2𝑥 5 + 𝑥 = 𝑣
...

1−𝑥 2

𝑑𝑦
Find
𝑑𝑥

• Solution
• Let 𝑢 =

𝑥2

+ 1 and 𝑣 = 1

• So, use quotient rule: 𝑦 =
•∴

𝑑𝑢
𝑑𝑥

= 2𝑥,

𝑑𝑣
𝑑𝑥

− 𝑥 2
...
𝑑𝑥−𝑢𝑑𝑥
𝑣2

= −2𝑥
...


Find

𝑑𝑦
𝑑𝑥

• Solution
• Let 𝑢 = 𝑥 2 − 2𝑥 3
...
Find

𝑑𝑦

...
5

= 𝑒 0
...
5𝑥

• Let f 𝑥 = −0
...

• So, use exponential function rule: 𝑦 = 𝑒 𝑓(𝑥) ⟹

•

𝑑𝑓
•
= 𝑓 ′ 𝑥 = −0
...
5𝑒 −0
...

𝑥

Find

𝑑𝑦
𝑑𝑥

• Solution
• 𝑦 = ln

1
𝑥

⟹ 𝑦 = ln

1
𝑥 0,5

= ln 𝑥 0,5

−1

= ln 𝑥 −0
...
5
• So, use logarithmic function rule: 𝑦 = ln 𝑓 𝑥 ⟹
𝑑𝑓
= 𝑓 ′ 𝑥 = −0
...
5𝑥 −1,5
0
...
5
𝑥

• Then
•⟹

𝑑𝑦
𝑑𝑥

=

𝑓′(𝑥)
𝑓(𝑥)

ECONOMIC APPLICATIONS – Ch 8
...
10
• Two frequently encountered applications derivatives are:
• Marginal revenue function
• Marginal cost function

MARGINAL REVENUE and MARGINAL COST, MC
• Recall that total revenue is 𝑅 = 𝑝𝑞
...
e
...
e
...
e
...

• Find marginal cost, 𝑀𝐶
...
5𝑞 + 100
...


SOLUTION, MC
• SOLUTION, MC

• 𝐶 = 3𝑞 3 − 54𝑞 2 + 500𝑞 + 2592
• Note that 𝑇𝑉𝐶 = 3𝑞 3 − 54𝑞 2 + 500𝑞, while TFC = 2592
• ∴ 𝑀𝐶 =

𝑑𝐶
𝑑𝑞

= 9𝑞 2 − 108𝑞 + 500

• Check for yourselves to see that MC =

𝑑𝑇𝑉𝐶
𝑑𝑞

• SOLUTION, MR

• 𝑝 = −0
...

𝑑𝑞

First have to find 𝑅

• 𝑅 = 𝑝𝑞 ⟹ 𝑅 = −0
...
5𝑞 2 + 100𝑞
• ∴ 𝑀𝑅 =

𝑑𝑅
𝑑𝑞

= −𝑞 + 100



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