7. The Product Rule

Loading...

Из курса от партнера University of London

Quantitative Foundations for International Business

164 оценки

University of London

Quantitative Foundations for International Business

164 оценки

Курс 3 из 6 — Specialization International Business Essentials

This course provides the essential mathematics required to succeed in the finance and economics related modules of the Global MBA, including equations, functions, derivatives, and matrices. You can test your understanding with quizzes and worksheets, while more advanced content will be available if you want to push yourself. This course forms part of a specialisation from the University of London designed to help you develop and build the essential business, academic, and cultural skills necessary to succeed in international business, or in further study. If completed successfully, your certificate from this specialisation can also be used as part of the application process for the University of London Global MBA programme, particularly for early career applicants. If you would like more information about the Global MBA, please visit https://mba.london.ac.uk/. This course is endorsed by CMI

Из урока

Derivatives

An important topic in many scientific disciplines- including economics- is the study of how quickly quantities change. The concept used to describe the rate of change of a function is known as the derivative. In this lecture, we will define the derivative of a function, and share some of the important rules for calculating it.

Познакомьтесь с преподавателями

George Kapetanios
Professor of Finance and Econometrics
King's College London

[MUSIC]

The correct rule for differentiating a product is

f(x) = f(x) x g(x).

We have that f prime (x)

= f prime (x) x g(x) +

f(x) x g prime (x).

Therefore, we have that the derivative of a product

of two functions is equal to the derivative of

the first term times the second function plus

the first times the derivative of the second.

In Leibniz's notation,

the product rule is expressed

as d/dx[f(x) x g (x)]

= [d/dx f(x)] x g(x)

+ f(x) x [d/dx g(x)].

If we use the product rule to find

h prime of x where h of x is equal to

(2x to the of power 6 + x top

the power of 2) (7x to the power

of 5- x to the power of 4).

We have h(x) =

f(x) x g(x) where f(x)

= (2x to the power of 6

+ x to the power of 2) and

g(x) = (7x to the power of

5- x to the power of 4).

The solution is f prime (x)

= (12x to the power of

5 + 2x) and g prime (x)

= (35x to the power of

4- 4x to the power of 3).

Thus, h prime (x) =

f prime (x) x g(x) +

f(x) x g prime (x).

Then we have (12x to the power

of 5 + 2x) x (7x to the power

of 5- x to the power of 4) +

(2x to the power of 6 + x to

the power of 2) x (35x to the power

of 4- 4x to the power of 3).

It is useful to simplify the answer by expanding in

order to obtain just one polynomial.

By multiplying each term,

we get, h prime (x) = 154 x

to the power of 10- 21x to

the power of 9 + 49x to the power

of 6- 4x to the power of 5.

For example, we can use the product rule for

differentiation in a simple economic framework.

Let's D(P) to denote the demand function for product.

By selling D(P) units, at price B per unit,

revenues are R(P) given by R(P) = PD (P).

From the economic theory, we know that normally, D prime (P) is negative.

This is true because the demand for goods goes down when the price increases.

According to the product rule for the differentiation,

R prime (P) = D(P) + PD prime (P),

which is the economic interpretation behind our derivative.

If for instance B increases by one pound,

the revenue R (P) will change for two reasons.

Firstly, R (P) increases by

1 x D(P) because each of the T(P)

will be worth one pound more.

However, the one pound more in the price

per unit will lead to change in the demand

by D(P + 1)- D(P) units,

which is about D prime (P).

The loss due to a one pound more in the price

per unit is therefore -PD prime P.

This must be subtracted from D (P) in order

to get R prime (P) as in our equation.

The resulting expression we represent the fact that R prime (P),

the total rate of change of R(P) is what

the seller earn less what he lose.

Let's differentiate a product of three factors.

If we have y = (x + 4)(x

+ 6)(x + 3),

y prime will be equal

to d/dx(x + 4)(x

+ 6)(x + 3) +

(x + 4) x d/dx (x

+ 6) (x + 3) +

(x + 4)(x + 6) x

d/dx (x + 3).

And now we get,

1(1)(x+6)(x+3) +

(x+4)(1) (x+3)

+ (x+4) (x+6)(1).

Then we get x to the power of

2 + 3x + 6x +18 + x

to the power of 2 + 4x +

3x + 12 + x to the power

of 2 + 6x + 4x + 24.

Which is equal to 3x to the power of 2 + 26x + 54.

[MUSIC]

Ознакомьтесь с нашим каталогом

Присоединяйтесь бесплатно и получайте персонализированные рекомендации, обновления и предложения.

Coursera делает лучшее в мире образование доступным каждому, предлагая онлайн-курсы от ведущих университетов и организаций.

© Coursera Inc., 2019 Все права защищены.

Загрузить из App Store

Загрузить в Google Play