Lecture 2 notes

Lecture Note 2

This is a lecture note originally written by a student and partly modified and certified by the

instructor.

CHAIN RULE

𝑑𝑦

𝑑𝑥 = 𝑑𝑦 𝑑𝑢 ∙

𝑑𝑢 𝑑𝑥

Example:

𝑦 = (𝑥3_{− 1)}7

Solution

Let 𝑥3− 1 = 𝑢

𝑑𝑢 𝑑𝑥 = 3𝑥2 𝑑𝑦 𝑑𝑢= 7𝑢6

𝑑𝑦

𝑑𝑢= 7(𝑥3− 1)6 𝑑𝑦

𝑑𝑥 = 7(𝑥3− 1)6∗ 3𝑥2

Example- Marginal Revenue Product

Rate of change with respect to number of employees

 A manufacturer determines that m employees will produce a total of q units per day where; 𝑞 =

(10𝑚2_{+ 1)}2_{. Let 𝑟 be revenue. If demand equation for the product is 𝑝 = 1000 − 5𝑞,}

determine _𝑑𝑚𝑑𝑟 (the rate of change of revenue with respect to change in number of employees also termed marginal revenue product).

Solution

Revenue = price × quantity

𝑟(𝑞) = (1000 − 5𝑞)𝑞 𝑟(𝑞) = 1000𝑞 − 5𝑞2

𝑑𝑟 𝑑𝑚=

𝑑𝑟 𝑑𝑞⋅

𝑑𝑞 𝑑𝑚

𝑑𝑟

𝑑𝑞= 1000 − 10(10𝑚 2_{+ 1)}2

o _𝑑𝑚𝑑𝑞 = 2(10𝑚2_{+ 1) ∗ 20𝑚 = 40𝑚(10𝑚}2_{+ 1)} 𝑑𝑟

𝑑𝑚= [1000 − 10(10𝑚2+ 1)2][40𝑚(10𝑚2+ 1)] 𝑑𝑟

𝑑𝑚= 40,000𝑚(10𝑚2+ 1) − 400(10𝑚2+ 1)3

THE GENERALIZED POWER FUNCTION (CHAIN RULE)

𝑦 = (𝑥𝑛_{+ 1)}𝑝

𝑑

𝑑𝑥(𝑢𝑎) = 𝑎𝑢𝑎−1⋅ 𝑑𝑢 𝑑𝑥

SPECIAL FUNCTIONS



𝑦 = 𝐼𝑛𝑥

𝑑

𝑑𝑥

(𝐼𝑛𝑢) =

1

𝑢

⋅

𝑑𝑢

𝑑𝑥



𝑦 = 𝑒

𝑥

𝑑

𝑑𝑥

(𝑒

𝑢

) = 𝑒

𝑢

⋅

𝑑𝑢

𝑑𝑥



𝑑

𝑑𝑥

(log

𝑏

𝑢) =

1 (𝐼𝑛𝑏)𝑢

⋅

𝑑𝑢 𝑑𝑥

Examples:

1.

𝑦 = 𝐼𝑛𝑥

Solution

let

𝑢 = 𝑥

𝑑𝑢

𝑑𝑥

= 1

𝑑𝑦 𝑑𝑥

=

1 𝑢

⋅

𝑑𝑢 𝑑𝑥

=

1 𝑥

⋅ 1

2.

𝑦 = 𝐼𝑛(𝑥

2

+ 5)

2

Solution

Let

𝑢 = (𝑥

2

+ 5)

2

𝑑𝑢

𝑑𝑦

𝑑𝑥

=

1

(𝑥

2

+ 5)

2

∗ 2(𝑥

2

+ 5)2𝑥

𝑑𝑦

𝑑𝑥

=

4𝑥

𝑥

2

+ 5

3.

𝑦 = 𝐼𝑛(𝑥

2

+ 2)

−1

Solution

𝑑𝑦

𝑑𝑥

=

1

(𝑥

2

+ 2)

−1

⋅ −1(𝑥

2

+ 2)

−2

2𝑥

𝑑𝑦

𝑑𝑥

= (𝑥

2

+ 2).

−2𝑥

(𝑥

2

+ 2)

2

𝑑𝑦

𝑑𝑥

=

−2𝑥

𝑥

2

+ 2

4.

𝑦 = 𝑒

𝑥

Solution

Let

𝑢 = 𝑥

𝑑𝑢

𝑑𝑥

= 1

𝑑𝑦

𝑑𝑥

= 𝑒

𝑢

⋅

𝑑𝑢

𝑑𝑥

= 𝑒

𝑥

⋅ 1

𝑑𝑦

𝑑𝑥

= 𝑒

𝑥

5.

𝑦 = 𝑒

(𝑥2+1)2

Solution

𝑑𝑦

𝑑𝑥

= 𝑒

(𝑥

2₊₁₎2

⋅ 2(𝑥

2

+ 1)2𝑥

𝑑𝑦

𝑑𝑥

= 𝑒

(𝑥

2₊₁₎2

⋅ 4𝑥(𝑥

2

+ 1)

Elasticity =

% 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦_{%𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑃𝑟𝑖𝑐𝑒}

Let

𝑄 = 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦

, and

𝑃 = 𝑃𝑟𝑖𝑐𝑒

% change in

𝑄 =

𝑑𝑄_𝑄

% change in P

=

𝑑𝑄_𝑄

Elasticity

=

𝑑𝑄/𝑄_{𝑑𝑃/𝑃}

=

𝑑𝑄_{𝑄 𝑑𝑃}𝑃

=

𝑑𝑄_𝑑𝑃𝑃_𝑄

Elasticity

=

𝑑𝑄_{𝑑𝑃𝑄(𝑝)}𝑃

E.g. The demand function for a product is

𝑞 = 𝑝

2

− 40𝑝 + 400

, where

𝑞 > 0

. Find;

a.

The point elasticity of demand

η(𝑝)

b.

Determine the point elasticity of demand when

𝑝 = 15.

Solution

a.

η(𝑝) =

𝑑𝑞_𝑑𝑝

⋅

_𝑞(𝑝)𝑝

𝑑𝑞

𝑑𝑝

= 2𝑝 − 40

η(𝑝) = 2𝑝 − 40 ⋅

𝑝

𝑝

2

− 40𝑝 + 400

η(𝑝) =

2𝑝

2

− 40𝑝

𝑝

2

− 40𝑝 + 400

b.

When

𝑝 = 15

η(𝑝) =

2(15)

2

− 40(15)

15

2

− 40(15) = 400

η(𝑝) = −6

RELATIVE EXTREMA

 A function 𝑓 is said to be ↑ on an interval 𝐼 when, for any two numbers 𝑥1, 𝑥2 in 𝐼, if 𝑥1< 𝑥2 ,

then𝑓(𝑥1) < 𝑓(𝑥2)

 A function 𝑓 is said to be ↓ on an interval 𝐼 when, for any two numbers 𝑥1, 𝑥2 in 𝐼, if 𝑥1< 𝑥2 ,

then𝑓(𝑥1) > 𝑓(𝑥2)

Rule 1:

Let 𝑓 be differentiable on the interval (𝑎, 𝑏) if, 𝑓′(𝑥) > 0 for all 𝑥 in (𝑎, 𝑏), then 𝑓 is increasing on (𝑎, 𝑏)

If, however, 𝑓′(𝑥) < 0 for all 𝑥 in (𝑎, 𝑏), then 𝑓 is decreasing on (𝑎, 𝑏)

Illustration

Given the function 𝑦 = 18𝑥 −2₃𝑥3. Determine whether the function is increasing or decreasing within the intervals (−2,1) and (4,7) on the 𝑥 axis.

𝑓′_{(𝑥) = 18 − 2𝑥}2

When 𝑥 = −2

𝑓′_{(−2) = 18 − 2(−2)}2_{= 10 > 0}

When 𝑥 = 1

𝑓′_{(1) = 18 − 2 = 16 > 0.}

Thus 𝑓 is ↑ on the interval 𝐼 = (−2,1) When 𝑥 = 4

𝑓′_{(4) = 18 − 2(4)}2 _{= −14 < 0}

When 𝑥 = 7

𝑓′_{(4) = 18 − 2(7)}2 _{= −80 < 0}

Thus 𝑓 is ↓ on the interval 𝐼 = (4,7)

In practice:

For this example, think of 𝑦 as revenue and

Relative Maxima and Relative Minima

Rule 2: Necessary condition for relative extrema

At the local (relative) minima or maxima, also called relative extremum, 𝑓′(𝑥) = 0.

Rule 2 however cannot tell exactly whether the point 𝑥 where 𝑓′(𝑥) = 0 is indeed a maximum or minimum. We need another rule to help us establish that.

Rule 3: Sufficiency (Second derivative) test for relative extrema

Suppose 𝑓′(𝑎) = 0

 If 𝑓′′(𝑎) < 0, then 𝑓 has a relative maximum at 𝑎.

 If 𝑓′′(𝑎) > 0, then 𝑓 has a relative minimum at 𝑎.

Illustration

a. find the points at which 𝑓(𝑥) is either minimum or maximum (i.e. points of the relative extrema).

b. Indicate whether a relative extremum from question is a relative maximum or minimum.

Solution.

a.

𝑓′_{(𝑥) = 4𝑥 − 4𝑥}3

𝑓′_{(𝑥) = 4𝑥(1 − 𝑥)(1 + 𝑥)}

𝑥 = 0, 𝑥 = 1, 𝑥 = −1

Thus, points 𝑥 = 0, 𝑥 = 1, and 𝑥 = −1 are relative extrema.

b. To establish whether 𝑓(𝑥) attains minimum or maximum values at these extrema, we apply rule 3.

𝑓′′_{(𝑥) = 4 − 12𝑥}2

At 𝑥 = 0, 𝑓′′(0) =4 − 0 = 4 > 0 → 𝑚𝑖𝑛𝑖𝑚𝑢𝑚

At 𝑥 = 1, 𝑓′′(1) = 4 − 12 = −8 < 0 → 𝑚𝑎𝑥𝑖𝑚𝑢𝑚

At 𝑥 = −1, 𝑓′′(−1) = 4 − 12 = −8 < 0 → 𝑚𝑎𝑥𝑖𝑚𝑢𝑚

Practical Example

Suppose price is given as 𝑝 =80−𝑞₄ , 0 ≤ 𝑞 ≤ 80, where 𝑞 is number of unit of products.

a. At what value of 𝑞 will maximum revenue be attained? b. What is this maximum revenue?

Solution:

Revenue= Price × Quantity

𝑟(𝑞) =(80 − 𝑞)

4 𝑞

𝑟(𝑞) = 80𝑞 − 𝑞2 4

At the quantity where maximum revenue is attained, 𝑟′(𝑞) = 0, and 𝑟′′(𝑞) < 0 a. 𝑟′(𝑞) = 80 − 2𝑞

80 − 2𝑞 = 0

Given the nature of the revenue function, then at quantity 𝑞 = 40, revenue could be minimum or maximum.

To establish whether revenue is maximum or minimum at 𝑞 = 40 we use the 2nd derivative test.

𝑟′′_{(𝑞) = −2 < 0}

Thus 𝑞 = 40 is a quantity level which generate the maximum revenue possible.

b.

Revenue

=

80(40)−402

4