Find derivative of: (3 - x^2)/(4 + x^2); Quotient Rule vs. Product Rule

In summary, the conversation discusses the incorrect use of the Product Rule method in finding the derivative of a function. The correct answer is obtained using the Quotient Rule, leading to a discussion about the cancellation of terms and the importance of understanding the principles behind mathematical operations. The expert reminds the individuals to review their work and to apply what they have learned in order to solve similar problems in the future.
  • #1
Lo.Lee.Ta.
217
0
Hi, I got the right answer when I used the Quotient Rule but not when I used the Product Rule...
I think it might be an algebra mistake...

Product Rule Method:

f'(x) = (3 - x^2)*(4 + x^2)^-1

= (3 - x^2)[(-1(4 + x^2)^-2)*2x] + [(4 + x^2)^-1](-2x)

= [(3 - x^2)(-2x)]/[(4 + x^2)^2] + [(-2x)/(4 + x^2)]

To get the same denominator here, I thought I might square the factor on the right.

= (-6x + 2x^3 + 4x^2)/[(4 + x^2)^2] <----- This is not right.


THE RIGHT ANSWER IS: -14x/[(4 + x^2)^2]
I found that by using the Quotient Rule.
...But where am I going wrong with the Product Rule...?


P.S. Oh, and a side note:
If I wrote out the Product Rule answer like this: (-6x + 2x^3 + 4x^2)/(x^4 + 8x^2 + 16)

Would the 4x^2 and the 8x^2 be able to cancel somewhat?
Sometimes I get confused about what can cancel.
Could it turn into:
(-6x + 2x^3)/(x^4 + 2 + 16) = (-6x + 2x^3)/(x^4 + 18)?

I know that's not the right answer, but is this how it would cancel?

Thank you very much!
 
Physics news on Phys.org
  • #2
given a fraction like (a+b / (a + c) you can't cancel the a terms out to get b / c

but if you have a * b / a * c then you can cancel the a factors out to get b / c

with that in mind try reviewing your work to see where you went wrong, the product and quotient rules should give the same answers.
 
  • #3
rewritten a bit more correctly what you surely meant
Lo.Lee.Ta. said:
Hi, I got the right answer when I used the Quotient Rule but not when I used the Product Rule...
I think it might be an algebra mistake...

Product Rule Method:

f(x) = (3 - x^2)*(4 + x^2)^-1

f'(x) = (3 - x^2)[(-1(4 + x^2)^-2)*2x] + [(4 + x^2)^-1](-2x)

= [(3 - x^2)(-2x)]/[(4 + x^2)^2] + [(-2x)/(4 + x^2)]

To get the same denominator here, I thought I might square the factor on the right.

= (-6x + 2x^3 + 4x^2)/[(4 + x^2)^2] <----- This is not right.
As far as I understand what you are saying, there is just no rationale for your operation of squaring. You have a sum, represent it X + Y . You are saying if you square a bit of it -> X + Y2 that will be the same thing?

You have something of form A/B2 + C/B .

In general how do you combine that into one?
 
  • #4
Lo.Lee.Ta. said:
Hi, I got the right answer when I used the Quotient Rule but not when I used the Product Rule...
I think it might be an algebra mistake...

Product Rule Method:

f'(x) = (3 - x^2)*(4 + x^2)^-1
Be careful to write what you mean! This is f, not f'. You have not yet differentiated.
[itex]f(x)= (3- x^2)(4+ x^2)^{-1}[/itex]

f'(x)= (3 - x^2)[(-1(4 + x^2)^-2)*2x] + [(4 + x^2)^-1](-2x)
Okay, this is correct.

= [(3 - x^2)(-2x)]/[(4 + x^2)^2] + [(-2x)/(4 + x^2)]

To get the same denominator here, I thought I might square the factor on the right.
No. Squaring a number changes the number. Multiply the numerator and denominator by [itex]4+ x^2[/itex]

= (-6x + 2x^3 + 4x^2)/[(4 + x^2)^2] <----- This is not right.
Yes, it is not right. If you have 2/3 and want to make the denominator 9, you multiply both numerator and denominator to get 6/9 which, of course, represents the same number as 2/3. If you square, you get 4/9 which is NOT the same.

Going back to
[tex]-\frac{2x(3- x^2)}{(4+x^2)^2}- \frac{2x}{4+ x^2}[/tex]
if you multipy both numerator and denominator of the second fraction by [itex]4+ x^2[/itex] you get
[tex]-\frac{6x- 2x^3}{(4+ x^2)^2}- \frac{8x+ 2x^3}{(4+ x^2)^2}[/tex]
[tex]= \frac{-6x+ 2x^3- 8x- 2x^3}{(4+ x^2)^2}= \frac{-14x}{(4+x^2)^2}[/tex]

THE RIGHT ANSWER IS: -14x/[(4 + x^2)^2]
I found that by using the Quotient Rule.
...But where am I going wrong with the Product Rule...?P.S. Oh, and a side note:
If I wrote out the Product Rule answer like this: (-6x + 2x^3 + 4x^2)/(x^4 + 8x^2 + 16)

Would the 4x^2 and the 8x^2 be able to cancel somewhat?
Sometimes I get confused about what can cancel.
Could it turn into:
(-6x + 2x^3)/(x^4 + 2 + 16) = (-6x + 2x^3)/(x^4 + 18)?

I know that's not the right answer, but is this how it would cancel?

Thank you very much!
 
Last edited by a moderator:
  • #5
WOW! HallsofIvy! You gave me the BEST answer I have ever gotten in my experience on Physics Forums!

You wrote out EXACTLY where I went wrong, and explained WHY it was wrong!

I get it now! My squaring method was off!

I worked the problem by multiplying instead by (4 + x^2) and got the right answer!

Thank you SO much for your help!
With the other answers, I didn't understand what was wrong with my method!

If this were YahooAnswers, you'd get voted "Best Answer!" :)
 
  • #6
Lo.Lee.Ta. said:
WOW! HallsofIvy! You gave me the BEST answer I have ever gotten in my experience on Physics Forums!

You wrote out EXACTLY where I went wrong, and explained WHY it was wrong!

I get it now! My squaring method was off!

I worked the problem by multiplying instead by (4 + x^2) and got the right answer!

Thank you SO much for your help!
With the other answers, I didn't understand what was wrong with my method!

If this were YahooAnswers, you'd get voted "Best Answer!" :)

Just apply what you've learned: [tex]\frac{(3-x^{2})'(4+x^{2})-(4+x^{2})'(3-x^{2})}{(4+x^{2})^{2}}=\frac{-14x}{(x^{2}+4)^{2}}[/tex]
 
  • #7
Lo.Lee.Ta. said:
WOW! HallsofIvy! You gave me the BEST answer I have ever gotten in my experience on Physics Forums!

You wrote out EXACTLY where I went wrong, and explained WHY it was wrong!

I get it now! My squaring method was off!

I worked the problem by multiplying instead by (4 + x^2) and got the right answer!

Thank you SO much for your help!
With the other answers, I didn't understand what was wrong with my method!

If this were YahooAnswers, you'd get voted "Best Answer!" :)

You heard it here first. :-p

epenguin said:
rewritten a bit more correctly what you surely meant

...

As far as I understand what you are saying, there is just no rationale for your operation of squaring. You have a sum, represent it X + Y . You are saying if you square a bit of it -> X + Y2 that will be the same thing?

You have something of form A/B2 + C/B .

In general how do you combine that into one?
 

Related to Find derivative of: (3 - x^2)/(4 + x^2); Quotient Rule vs. Product Rule

1. What is the Quotient Rule and how is it used to find the derivative of a function?

The Quotient Rule is a differentiation rule that is used to find the derivative of a function that is expressed as the quotient of two other functions. The formula for the Quotient Rule is (f'(x)g(x) - g'(x)f(x)) / (g(x))^2, where f(x) and g(x) are the two functions in the quotient. To use the Quotient Rule, you must first find the derivatives of each of the two functions, then plug them into the formula to get the derivative of the original function.

2. How does the Product Rule differ from the Quotient Rule?

The Product Rule is another differentiation rule that is used to find the derivative of a function that is expressed as the product of two other functions. The formula for the Product Rule is f'(x)g(x) + g'(x)f(x), where f(x) and g(x) are the two functions in the product. The main difference between the Product Rule and the Quotient Rule is that the Product Rule is used for products of functions, while the Quotient Rule is used for quotients of functions.

3. When should I use the Quotient Rule to find the derivative of a function?

You should use the Quotient Rule when you have a function that is expressed as the quotient of two other functions. This could be a function in the form of (f(x))/g(x), where f(x) and g(x) are two separate functions. The Quotient Rule is also useful when the two functions in the quotient are more complex and cannot be easily differentiated using the other basic rules (such as the Power Rule or the Chain Rule).

4. Can the Quotient Rule and the Product Rule be used together to find the derivative of a function?

Yes, the Quotient Rule and the Product Rule can be used together to find the derivative of a function. In some cases, a function may require both rules to be applied in order to find the derivative. For example, if a function is expressed as the product of two other functions, and one of those functions is also a quotient of two functions, then both the Product Rule and the Quotient Rule would need to be used to find the derivative.

5. Are there any alternative methods to finding the derivative of a function besides using the Quotient Rule and the Product Rule?

Yes, there are other methods for finding the derivative of a function. Some common alternatives include using the Power Rule, the Chain Rule, and the Sum/Difference Rule. The choice of which method to use depends on the complexity of the function and which rule will make the process of finding the derivative easier and more efficient.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
696
  • Calculus and Beyond Homework Help
Replies
25
Views
693
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
883
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
642
  • Calculus and Beyond Homework Help
Replies
7
Views
725
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
2K
Back
Top