- #1
UrbanXrisis
- 1,192
- 1
what is the derivative of 1/(4x^2+3x-x)
I got 1/4x is that correct?
I got 1/4x is that correct?
Derivative of 1/(4x^2 + 3x - x)
- Thread starter UrbanXrisis
- Start date
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- Derivative
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Homework Help Overview
The discussion revolves around finding the derivative of the function 1/(4x^2 + 3x - x). Participants are exploring the correct application of differentiation rules in the context of rational functions.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning
Approaches and Questions Raised
- Participants are questioning the validity of different approaches to finding the derivative, including the quotient rule and the chain rule. There is also a discussion about the incorrect assumption that the sum of fractions can be split in a certain way.
Discussion Status
Some participants have provided guidance on using the quotient rule or rewriting the function for differentiation. There is an acknowledgment of previous misunderstandings, and the conversation is focused on clarifying the correct methods for differentiation.
Contextual Notes
Participants are addressing the misconception that the sum of fractions can be simplified incorrectly, highlighting the importance of understanding the rules of algebra in the context of calculus.
- #2
Doc Al
Mentor
- 45,584
- 2,460
How did you get that?
- #3
UrbanXrisis
- 1,192
- 1
1/(4x^2+3x-x) =
(1/4x^2)+(1/3x)-(1/x)=
((4x^2)^-1 )+ ((3x)^-1) - ((x)^-1))=
4x+3-1/x
derivative = 4x^-1 = 1/4x
(1/4x^2)+(1/3x)-(1/x)=
((4x^2)^-1 )+ ((3x)^-1) - ((x)^-1))=
4x+3-1/x
derivative = 4x^-1 = 1/4x
- #4
Ebolamonk3y
- 180
- 0
Wrong.
The answer is -(4x+1)/((2x^2)*(2x+1)^2)
The answer is -(4x+1)/((2x^2)*(2x+1)^2)
- #5
Doc Al
Mentor
- 45,584
- 2,460
1/(a+b) does not equal 1/a + 1/b !
If that were true, then this would follow:
[tex]\frac{1}{2} = \frac{1}{1+1} = \frac{1}{1} + \frac{1}{1} = 2[/tex]
Yikes!UrbanXrisis said:
If that were true, then this would follow:
[tex]\frac{1}{2} = \frac{1}{1+1} = \frac{1}{1} + \frac{1}{1} = 2[/tex]
- #6
UrbanXrisis
- 1,192
- 1
oh wow, what was I thinking? whooops. How would I solve this then?
- #7
faust9
- 690
- 2
You could use the quotient rule, or simply rewrite the equation as [itex](4x^2+2x)^{-1}[/itex] and use the regular chain rule.
PS: [itex]\frac{1}{A+B+C}[/itex] does not equal [itex]\frac{1}{A}+\frac{1}{B}+\frac{1}{C}[/itex] ; however, [itex]\frac{A+B+C}{D}[/itex] does equal [itex]\frac{A}{D}+\frac{B}{D}+\frac{C}{D}[/itex]. See the difference? You can't simply split a denominator but you can split up a numerator. Use partial fractions to split the denominator.
[edit] fixed my tex tags
PS: [itex]\frac{1}{A+B+C}[/itex] does not equal [itex]\frac{1}{A}+\frac{1}{B}+\frac{1}{C}[/itex] ; however, [itex]\frac{A+B+C}{D}[/itex] does equal [itex]\frac{A}{D}+\frac{B}{D}+\frac{C}{D}[/itex]. See the difference? You can't simply split a denominator but you can split up a numerator. Use partial fractions to split the denominator.
[edit] fixed my tex tags
Last edited:
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