Understanding the Product Rule: A Closer Look at Solving y=2x(1-x)^2

In summary, the book introduces the quotient rule for solving equations of the form y=x^3 instead of the product rule.
  • #1
fitz_calc
41
0

Homework Statement


y=2x(1-x)^2


Homework Equations





The Attempt at a Solution



y=2x(1-x)^2

y`=-x
-----------------
(1-x)^1/2

I thought i was done here. The book takes a few more steps. it adds:

-x
-----------------
(1-x)^1/2

to this:

2(1-x)^1/2 (1-x)^1/2
------------ x -----------
1 (1-x)^1/2


Why does the last step exist?
 
Physics news on Phys.org
  • #2
and one more question. i just got to a problem with sqrt(x+1) in the denominator. my origional thought was to put it in the numerator to the power of -1/2 and use the product rule.

my book uses the quotient rule instead. how do i know which one is the best one to use?
 
Last edited:
  • #3
fitz_calc said:

Homework Statement


y=2x(1-x)^2

Homework Equations


The Attempt at a Solution



y=2x(1-x)^2

y`=-x
-----------------
(1-x)^1/2

I thought i was done here. The book takes a few more steps. it adds:

-x
-----------------
(1-x)^1/2

to this:

2(1-x)^1/2 (1-x)^1/2
------------ x -----------
1 (1-x)^1/2Why does the last step exist?

Are the dotted lines supposed to indicate that ir is a fraction?

And have you used the chain rule yet?

What is the original function now?
Is it [tex]y=2x(1-x)^2[/tex] ?
 
Last edited:
  • #4
Did you do the product rule right?

f'(x)g(x) + f(x)g'(x)?

Or secondly, you can follow the definition of the limit and right

lim as x approaches deltax 2(x+dx)(1-(x+dx)^2 all over dx

where dx = delta x

Use that to confirm if yo did the product rule right, I can't understand the way you wrote your work. Looking at your original function, I don't understand how you got a fraction.
 
Last edited:
  • #5
PowerIso said:
Did you do the product rule right?

f'(x)g(x) + f(x)g'(x)?

PowerIso, can you understand what he's writing? If the original function is just [tex]y=2x(1-x)^2[/tex]

I see no need for the product rule as this is just a polynomial of the third degree...Power Rule.

Casey

Edit: I saw your edit...the fraction is messing me up too..that is why I believe he maybe copied the original function wrong...maybe he meant something else.:confused:
 
Last edited:
  • #6
Yes I can understand that, but his question regarded the product rule, so I tackled this problem with that view.
 

1. What is the product rule and how does it work?

The product rule is a mathematical rule used in calculus to find the derivative of a product of two functions. It is written as (fg)' = f'g + fg', where f' and g' represent the derivatives of the functions f and g, respectively. It is used when finding the rate of change of a quantity that is the product of two other quantities.

2. Can you give an example of a problem that uses the product rule?

One example of a problem that uses the product rule is finding the derivative of the function f(x) = x^2 * sin(x). Using the product rule, we can find that f'(x) = 2x * sin(x) + x^2 * cos(x).

3. How is the product rule different from the chain rule?

The product rule is used to find the derivative of a product of two functions, while the chain rule is used to find the derivative of a composition of two functions. In other words, the product rule is used when two functions are being multiplied together, while the chain rule is used when one function is being applied to another function.

4. Are there any special cases where the product rule does not apply?

Yes, there are special cases where the product rule does not apply. One example is when one of the functions is a constant. In this case, the derivative of the product can be found using the constant multiple rule instead of the product rule.

5. How can the product rule be applied in real-world situations?

The product rule can be applied in many real-world situations, such as in economics when calculating the marginal cost of producing two goods at the same time, or in physics when finding the rate of change of the position of an object that is accelerating while also experiencing a force. It is a useful tool for understanding and solving problems involving changing quantities.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
532
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
837
  • Calculus and Beyond Homework Help
Replies
7
Views
691
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
166
  • Calculus and Beyond Homework Help
Replies
18
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
704
  • Calculus and Beyond Homework Help
Replies
1
Views
961
Back
Top