lecture 1 business math

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Rate of Change and the Derivative

When two (or more ) variables say 𝑥 and 𝑦 are related to each other by a function

y = 𝑓(𝑥), there is always an interest in changes in one variable to a change in the other variable.

When the variable 𝑥 changes from 𝑥₀ to 𝑥₁, the change is measured by the difference 𝑥₁ − 𝑥₀. Using the symbol ∆ to denote the change, then:

∆𝑥 = 𝑥₁ − 𝑥₀

We use 𝑓(𝑥_𝑖) to denote the value of the function 𝑓(𝑥) when 𝑥 = 𝑥_𝑖

e.g. Given 𝑓 𝑥 = 5 + 𝑥2,

𝑓 0 = 5 + 02 = 5 and 𝑓 2 = 5 + 22 = 9.

When 𝑥 changes from an initial value of 𝑥₀ to a new value (𝑥₀ + ∆𝑥), the function

𝑦 = 𝑓(𝑥) also changes from 𝑓(𝑥₀) to 𝑓(𝑥₀ + ∆𝑥).

The change in 𝑦 per unit of change in 𝑥 is represented by the difference quotient

∆𝑦 ∆𝑥 =

𝑓 𝑥₀ + ∆𝑥 − 𝑓(𝑥₀) ∆𝑥

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Rate of Change and the Derivative

Example:

Given 𝑦 = 3𝑥2 − 6,

𝑓 𝑥₀ = 3 𝑥₀ 2 − 6 , and 𝑓 𝑥₀ + ∆𝑥 = 3 𝑥₀ + ∆𝑥 2 − 6 ∆𝑦

∆𝑥 =

𝑓 𝑥₀ + ∆𝑥 − 𝑓(𝑥₀)

∆𝑥 =

3 𝑥₀ + ∆𝑥 2 − 6 − (3 𝑥₀ 2 − 6) ∆𝑥

= 6𝑥0∆𝑥 + 3 ∆𝑥

2

∆𝑥 = 6𝑥0 + 3∆𝑥

∆𝑦

∆𝑥 can be evaluated for known values of 𝑥0 and ∆𝑥. For example, for 𝑥₀ = 5 and ∆𝑥 = 2, ∆𝑦

∆𝑥 = 6 ∗ 5 + 3 ∗ 2 = 36.

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The Derivative

Most times, we are interested in the rate of change of 𝑦 when ∆𝑥 is very small. In such cases, it is possible to obtain ∆𝑦

∆𝑥 by dropping all terms of ∆𝑥 in the difference quotient formula.

For example, in the last example, if ∆𝑥 is very small, then ∆𝑦

∆𝑥 ≈ 6 ∗ 5 = 30 The smaller the value of ∆𝑥 , the closer is the approximation of ∆𝑦

∆𝑥 to the true ∆𝑦 ∆𝑥. As ∆𝑥 approaches zero (meaning that it get closer and closer to, but never really reaches zero) 6𝑥₀ + 3∆𝑥 will approach 6𝑥₀.

Symbolically, this statement is expressed as ∆𝑦

∆𝑥 → 0 or ∆𝑥 → 0 or by the equation

lim

∆𝑥→0

∆𝑦

∆𝑥 = lim∆𝑥→06𝑥0 + 3∆𝑥 = 6𝑥0 The symbol lim

∆𝑥→0 is read: “The limit of ….. as ∆𝑥 approaches 0. If as ∆𝑥 → 0, the limit of the difference quotient ∆𝑦

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Derivative of a function 𝒇

Slope of line 𝑃𝑄 = 𝑚_𝑃𝑄 = 𝑓 𝑧 −𝑓(𝑎)

𝑧−𝑎 Let 𝑍 = 𝑎 + 𝑕

Then 𝑚_𝑃𝑄 = 𝑓 𝑧 −𝑓(𝑎)

𝑕

As 𝑄 moves along the curve toward 𝑃, 𝑧 approaches 𝑎. This also means 𝑕 → 0

The slope of a tangent line at point (𝑎, 𝑓 𝑎 ) is given as the limiting value of the slope:

𝑚_𝑡𝑎𝑛 = lim

𝑧→𝑎

𝑓 𝑧 − 𝑓(𝑎)

𝑧 − 𝑎 = lim𝑕→0

𝑓 𝑎 + 𝑕 − 𝑓(𝑎) 𝑕

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Derivative of a function 𝒇

The derivative of a function 𝑓 is the function denoted by 𝑓′ which is defined by:

lim

𝑧→𝑥

𝑓(𝑧) − 𝑓(𝑥)

𝑧 − 𝑥 = lim𝑕→0

𝑓(𝑥 + 𝑕) − 𝑓(𝑥)

𝑕 = 𝑓′(𝑥)

provided the limit exist.

If 𝑓′(𝑎) can be found, the original function 𝑓 is said to be differentiable at

𝑎, and 𝑓′(𝑎) is called the derivative of 𝑓 at 𝑎 or the derivative of 𝑓 with respect to 𝑥 at 𝑎.

The process of finding the derivative is called Differentiation.

Example:

If 𝑓 𝑥 = 𝑥2, find the derivative of 𝑓 𝑓′ 𝑥 = lim

𝑕→0

𝑓(𝑥 + 𝑕) − 𝑓(𝑥)

𝑧 − 𝑥 = lim𝑕→0

𝑥 + 𝑕 2 − 𝑥2 𝑕

= lim

𝑕→0

𝑥2 + 2𝑥𝑕 + 𝑕2 − 𝑥2 𝑕

𝑓′ 𝑥 = lim

𝑕→0

2𝑥𝑕 + 𝑕2

𝑕 = 𝑓

′ _{𝑥 = lim} 𝑕→0

𝑕(2𝑥 + 𝑕)

𝑕 = lim𝑕→0(2𝑥 + 𝑕) = 2𝑥

Common Notations of a derivative of a function: 𝑑𝑦

𝑑𝑥, 𝑑

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Rules of Differentiation

1. Constant Function Rule

The derivative of a constant 𝑦 = 𝑓 𝑥 = 𝑘 is zero. 𝑑𝑦

𝑑𝑥 = 0 or 𝑑𝑘

𝑑𝑥 = 0 or 𝑓

′ _{𝑥 = 0}

Other expressions as below are correct

𝑑

𝑑𝑥𝑦 = 𝑑

𝑑𝑥𝑓 𝑥 = 𝑑

𝑑𝑥𝑘 = 0

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2. Power-Function Rule

The derivative of a power function 𝑦 = 𝑓 𝑥 = 𝑥𝑛

𝑑𝑦

𝑑𝑥 = 𝑓

′ _{𝑥 =} 𝑑

𝑑𝑥𝑥

𝑛 _{= 𝑛𝑥}𝑛−1 Example: Find the derivative of 𝑦 = 𝑥4

𝑑𝑦

𝑑𝑥 = 4𝑥

4−1 _{= 4𝑥}3 Example: Find the derivative of 𝑦 = 1

𝑥4 Note that 𝑦 = 1

𝑥4 = 𝑥 −4_. Therefore, 𝑑𝑦

𝑑𝑥 = −4𝑥

−4−1 _{= −4𝑥}−5 _{= −} 4 𝑥5

Note that since the derivative is a function of 𝑥, different values of 𝑥 may result in different values for the derivative.

Example: 𝑓′ 𝑥4 = 4𝑥3

At 𝑥 = 1, 𝑓′ 1 = 4.

At 𝑥 = 2, 𝑓′ 2 = 4 ∗ 23 = 32

3. Generalized Power-Function Rule

For a function 𝑓 = 𝑐𝑥𝑛

The derivative 𝑑𝑦

𝑑𝑥 = 𝑐𝑛𝑥 𝑛−1 Example: 𝑓 = 5𝑥3

𝑓′ 𝑥 = 5 ∗ 3𝑥2 = 15𝑥2

Example: 𝑦 = 4𝑥−3 𝑑𝑦

𝑑𝑥 = 4 ∗ −3𝑥

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Rules of Differentiation Involving two or more functions of the same variable

1. Sum-Difference Rule

The derivative of a function 𝑦 = 𝑓 𝑥 ± 𝑔(𝑥)

𝑑

𝑑𝑥 𝑓 𝑥 ± 𝑔 𝑥 = 𝑑

𝑑𝑥𝑓 𝑥 ± 𝑑

𝑑𝑥𝑔 𝑥 = 𝑓′ 𝑥 + 𝑔′(𝑥)

Example: Find the derivative of the function

𝑦 = 3𝑥3 + 2𝑥2 + 𝑥 𝑑𝑦

𝑑𝑥 = 𝑑

𝑑𝑥 3𝑥

3 ₊ 𝑑

𝑑𝑥 2𝑥

2 ₊ 𝑑

𝑑𝑥 𝑥 𝑑𝑦

𝑑𝑥 = 9𝑥

2 _{+ 4𝑥 + 1}

Example: Find the derivative of the total-cost function

𝐶 = 𝑄3 − 4𝑄2 + 10𝑄 + 75.

𝑑𝐶

𝑑𝑄 = 3𝑄

2 _{− 8𝑄 + 10}

2. Product Rule

Given the function 𝑦 = 𝑓 𝑥 ∗ 𝑔(𝑥), the derivative of 𝑦 is expressed as:

𝑑

𝑑𝑥 𝑓 𝑥 𝑔 𝑥 = 𝑓 𝑥 𝑑

𝑑𝑥𝑔(𝑥) + 𝑔(𝑥) 𝑑

𝑑𝑥𝑓 𝑥 =

𝑓 𝑥 𝑔′ 𝑥 + 𝑔 𝑥 𝑓′(𝑥)

Proof:

𝐹′ 𝑥 = lim

𝑕→0

𝑓 𝑥 + 𝑕 − 𝑓(𝑥)

𝑕 = lim𝑕→0

𝑓 𝑥 + 𝑕 𝑔 𝑥 + 𝑕 − 𝑓 𝑥 𝑔(𝑥) 𝑕

Add and subtract 𝑓 𝑥 𝑔(𝑥 + 𝑕)

lim 𝑕→0

𝑓 𝑥 + 𝑕 𝑔 𝑥 + 𝑕 − 𝑓 𝑥 𝑔 𝑥 + 𝑓 𝑥 𝑔 𝑥 + 𝑕 − 𝑓 𝑥 𝑔(𝑥 + 𝑕) 𝑕

Regroup

lim

𝑕→0

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Rules of Differentiation Involving two or more functions of the same variable

lim

𝑕→0

𝑓 𝑥+𝑕 −𝑓 𝑥 𝑔 𝑥+𝑕

𝑕 + lim_𝑕→0

𝑓 𝑥 [𝑔 𝑥+𝑕 −𝑔 𝑥 ]

𝑕

lim

𝑕→0

𝑓 𝑥+𝑕 −𝑓 𝑥

𝑕 ∗ lim_𝑕→0

𝑔 𝑥+𝑕

𝑕 + lim_𝑕→0

[𝑔 𝑥+𝑕 −𝑔 𝑥 ]

𝑕 ∗ lim_𝑕→0 𝑓 𝑥

𝑕 If 𝐹 is differentiable, then 𝐹 is continuous, ⇒ lim

𝑕→0

𝑔 𝑥+𝑕

𝑕 =

𝑔(𝑥) and so

𝐹′ 𝑥 = 𝑓′ 𝑥 𝑔 𝑥 + 𝑓 𝑥 𝑔′(𝑥)

Example: Find the differential of the function F 𝑥 = (𝑥3 + 3𝑥)(4𝑥 + 5)

Let 𝑓 𝑥 = 𝑥3 + 3𝑥 and 𝑔 𝑥 = 4𝑥 + 5. Then,

𝑓′ 𝑥 = 3𝑥2 + 3 and 𝑔′ 𝑥 = 4

𝐹′ 𝑥 = 𝑥3 + 3𝑥 ∗ 4 + 4𝑥 + 5 ∗ 3𝑥2 + 3

𝐹′ 𝑥 = 4𝑥3 + 12𝑥 + 12𝑥3 + 15𝑥2 + 12𝑥 + 15

𝐹′ 𝑥 = 16𝑥3 + 15𝑥2 + 24𝑥 + 15 Example:

Find 𝑑𝑦

𝑑𝑥 if 𝑦 = 𝑥

2

3 + 3 𝑥− 1

3 + 5𝑥

Ans: 25 3 𝑥

2 3 + 1

3𝑥

−2₃ _{− 𝑥}−4₃ _{+ 15}

Example: Given 𝑦 = (𝑥 + 2)(𝑥 + 3)(𝑥 + 4), what is 𝑦′

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Rules of Differentiation

The Quotient Rule 𝑑

𝑑𝑥

𝑓(𝑥) 𝑔(𝑥) =

𝑔 𝑥 𝑓′ 𝑥 −𝑓 𝑥 𝑔′(𝑥) 𝑔 𝑥 2 Proof:

If 𝐹 𝑥 = 𝑓 𝑥 /𝑔(𝑥), then 𝐹 𝑥 𝑔 𝑥 = 𝑓(𝑥)

By the product rule, 𝑓′ 𝑥 = 𝐹′ 𝑥 𝑔 𝑥 + 𝑔′ 𝑥 𝐹 𝑥

Rearranging,

𝐹′ 𝑥 = 𝑓

′ _{𝑥 − 𝑔}′ _{𝑥 𝐹(𝑥)}

𝑔(𝑥)

But 𝐹 𝑥 = 𝑓(𝑥)/𝑔(𝑥) ⇒

𝐹′ 𝑥 = 𝑓

′ _{𝑥 −}𝑔′ 𝑥 𝑓 𝑥 𝑔 𝑥

𝑔(𝑥) =

𝑓′ 𝑥 𝑔 𝑥 −𝑓 𝑥 𝑔′(𝑥) 𝑔 𝑥 2

Example: Given 𝐹 𝑥 = 4𝑥2+3

2𝑥−1 , find 𝐹′(𝑥) Example: Marginal Revenue

If 𝑝 = 1000

𝑞+5 Revenue

(r) = 1000

𝑞+5 𝑞 =

1000 𝑞+5 𝑑𝑟

𝑑𝑞 =

5000

𝑞+5 2 At 𝑞 = 45, 𝑑𝑟

𝑑𝑞 = 2

Relationship between Marginal cost and Average-cost

Suppose Total-cost function is 𝐶 = 𝐶(𝑄)

Then Average-cost function is: 𝐴𝐶 = 𝐶 𝑄

𝑄 for 𝑄 > 0

The rate of change of 𝐴C wrt 𝑄 is given as: 𝑑

𝑑𝑄 𝐶 𝑄

𝑄 =

[𝐶′ 𝑄 ∗𝑄−𝐶 𝑄 ∗1]

𝑄2 =

1 𝑄[𝐶

′ _{𝑄 −} 𝐶 𝑄 𝑄 ] Then 𝑑 𝑑𝑄 𝐶 𝑄

𝑄 ≥ 0 iff 𝐶

′ _{𝑄 ≥} 𝐶 𝑄 𝑄 𝑑

𝑑𝑄 𝐶 𝑄

𝑄 ≤ 0 iff 𝐶

′ _{𝑄 ≤} 𝐶 𝑄 𝑄

Since the derivative 𝐶′(𝑄) is the marginal-cost function, and 𝐶 𝑄

𝑄 is the Average-cost function, the economic meaning of the two iff relationship is that:

The slope of the AC curve will be positive, zero, or negative if and only if the

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Rules of Differentiation

0 20 40 60 80 100 120 140

1 2 3 4 5 6 7 8 9 10

Chart Title

MC AC

𝑀𝐶 = 3𝑄2 − 24𝑄 + 60

A𝐶 = 𝑄2 − 12𝑄 + 60

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Rules of Differentiation

The Quotient Rule 𝑑

𝑑𝑥

𝑓(𝑥) 𝑔(𝑥) =

𝑔 𝑥 𝑓′ 𝑥 −𝑓 𝑥 𝑔′(𝑥) 𝑔 𝑥 2 Proof:

If 𝐹 𝑥 = 𝑓 𝑥 /𝑔(𝑥), then 𝐹 𝑥 𝑔 𝑥 = 𝑓(𝑥)

By the product rule, 𝑓′ 𝑥 = 𝐹′ 𝑥 𝑔 𝑥 + 𝑔′ 𝑥 𝐹 𝑥

Rearranging,

𝐹′ 𝑥 = 𝑓

′ _{𝑥 − 𝑔}′ _{𝑥 𝐹(𝑥)}

𝑔(𝑥)

But 𝐹 𝑥 = 𝑓(𝑥)/𝑔(𝑥) ⇒

𝐹′ 𝑥 = 𝑓

′ _{𝑥 −}𝑔′ 𝑥 𝑓 𝑥 𝑔 𝑥

𝑔(𝑥) =

𝑓′ 𝑥 𝑔 𝑥 −𝑓 𝑥 𝑔′(𝑥) 𝑔 𝑥 2

Example: Given 𝐹 𝑥 = 4𝑥2+3

2𝑥−1 , find 𝐹′(𝑥) Example: Marginal Revenue

If 𝑝 = 1000

𝑞+5 Revenue

(r) = 1000

𝑞+5 𝑞 =

1000 𝑞+5 𝑑𝑟

𝑑𝑞 =

5000

𝑞+5 2 At 𝑞 = 45, 𝑑𝑟

𝑑𝑞 = 2

The consumption Function

The consumption function expresses a relationship between the total national income 𝐼 and the total national

consumption 𝐶. That is: 𝐶 = 𝑓(𝐼)

The marginal propensity to consume, is defined as the rate of change of

consumption with respect to income. i.e. derivative of 𝐶 wrt 𝐼.

Marginal propensity to consume = 𝑑𝐶/𝑑𝐼

If savings 𝑆 is the difference between income 𝐼 and consumption 𝐶, then

𝑆 = 𝐼 − 𝐶

and

The derivative of 𝑆 wrt to 𝐼 is the marginal propensity to save, which is expressed as:

𝑑𝑆 𝑑𝐼 =

𝑑 𝐼 𝑑𝐼 −

𝑑 𝐶

𝑑𝐼 = 1 − 𝑑𝐶

𝑑𝐼

Example:

If the consumption function is 𝐶 =

5(2 𝐼3+3)

𝐼+10 , determine the marginal