# Calculus made easy by Thompson: exercise vii: 1

• tridianprime
In summary, the problem involves finding dw/dx using the chain rule. The derivatives of u, v, and w are calculated to be v^(-2), 6u+3, and 3/2x^2, respectively. However, the multiplication of these derivatives results in an incorrect answer. The correct answer is -(3x^2(3+3x^3))/(27((1/2)x^3 + (1/4)x^6)^3).
tridianprime
Hello, I require some assistance with this problem. It reads:

If, u=1/2x^3; v=3(u+u^2); w=1/v^2;

Find dw/dx.

It's clear that you should use the chain rule so I begin by calculating the derivative for each one.

I get v^(-2); 6u+3; and 3/2x^2.

I then have to multiply them but this is where the problem comes. The answer I derived was (3x^(2)-12u-6)/2v^3

This doesn't seem right and the answer say it isn't at the back of book.

Is the problem the multiplication?

tridianprime said:
If, u=1/2x^3; v=3(u+u^2); w=1/v^2;

Find dw/dx.

It's clear that you should use the chain rule so I begin by calculating the derivative for each one.

I get v^(-2); 6u+3; and 3/2x^2.

v-2 = what?
6u+3 = what?
etc.

In order of appearance: dw/dv then dv/du then du/dx.

tridianprime said:
Hello, I require some assistance with this problem. It reads:

If, u=1/2x^3; v=3(u+u^2); w=1/v^2;

Find dw/dx.

It's clear that you should use the chain rule so I begin by calculating the derivative for each one.

I get v^(-2); 6u+3; and 3/2x^2.
The first is incorrect. $w= 1/v^2= v^{-2}$ itself. The derivative is $w'= -2v^{-3}$
but apparently that is a typo since you have $v^3$ below.
(u= (1/2)x^3, not 1/(2x^3)?)

I then have to multiply them but this is where the problem comes. The answer I derived was (3x^(2)-12u-6)/2v^3
Specifically, you have $(-2v^{-3})(6u+ 3)(3/2)x^2$. With [/itex]u= (1/2)x^3[/itex], $6u+ 3= 3x^3+ 3$ so this is -9(x^5+ 3x^2)/v^3

This doesn't seem right and the answer say it isn't at the back of book.

Is the problem the multiplication?
What is the answer in the book? Do they have x, u, and v or have the converted everything to x?

Everything is x.

It reads: -(3x^2(3+3x^3))/(27((1/2)x^3 + (1/4)x^6)^3)

## 1. What is the purpose of exercise vii: 1 in "Calculus made easy by Thompson"?

Exercise vii: 1 in "Calculus made easy by Thompson" is meant to help readers practice and reinforce their understanding of the concepts covered in the previous chapters. It includes a variety of problems that cover different aspects of calculus, allowing readers to test their knowledge and skills.

## 2. Is exercise vii: 1 suitable for beginners in calculus?

Yes, exercise vii: 1 in "Calculus made easy by Thompson" is suitable for beginners in calculus. It starts with relatively simple problems and gradually increases in difficulty, making it accessible for those who are new to the subject.

## 3. How can exercise vii: 1 help improve my understanding of calculus?

Exercise vii: 1 in "Calculus made easy by Thompson" challenges readers to apply their knowledge and skills in a practical way. By attempting and solving the problems, readers can identify any gaps in their understanding and work on improving those areas.

## 4. Can I use exercise vii: 1 as a study tool for exams?

Yes, exercise vii: 1 can be a helpful study tool for exams. It covers a wide range of topics in calculus and provides practice problems that are similar to those that may appear on exams. By regularly practicing with exercise vii: 1, readers can build their confidence and improve their performance on exams.

## 5. Are the solutions to exercise vii: 1 available in the book?

Yes, the solutions to exercise vii: 1 are provided in the book "Calculus made easy by Thompson". They are located at the end of the chapter, allowing readers to check their work and understand the steps and methods used to solve each problem.

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