Is it okay to treat dy/dx as a fraction when solving for derivatives?

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Homework Help Overview

The discussion revolves around the treatment of dy/dx as a fraction in the context of derivatives. Participants explore whether it is appropriate to manipulate dy/dx in fractional form and the implications of doing so in various scenarios.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the simplification of expressions involving dy/dx and question the handling of signs in calculations. There is also exploration of the relationship between dy/dx and dx/dy, with some participants expressing uncertainty about the implications of this interchange.

Discussion Status

The conversation includes attempts to clarify the treatment of dy/dx and its manipulation. Some participants provide affirmations regarding the validity of treating dy/dx as a fraction, while others seek confirmation of their reasoning and the correctness of their statements.

Contextual Notes

Participants express concern about adhering to rules related to derivatives and the proper treatment of notation, indicating a desire to ensure their understanding aligns with established conventions.

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Homework Statement


Am I right to treat dy/dx as a fraction in this scenerio?

Homework Equations





The Attempt at a Solution




[tex]= \frac{\frac{-dy}{dx}} {(\frac{dy}{dx})^2}[/tex]
[tex]= \frac{\frac{-dy}{dx}} {\frac{dy}{dx} (\frac{dy}{dx})}[/tex]
[tex]= \frac{1}{\frac{dy}{dx}}[/tex]

Also, would we be able to multiple out dy/dx if we had for example:

[tex]2 = \frac{1}{\frac{dy}{dx}}[/tex]
[tex]2(\frac{dy}{dx}) = 1[/tex]
 
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Nano-Passion said:

Homework Statement


Am I right to treat dy/dx as a fraction in this scenerio?
Sure.
Nano-Passion said:

Homework Equations





The Attempt at a Solution




[tex]= \frac{\frac{-dy}{dx}} {(\frac{dy}{dx})^2}[/tex]
[tex]= \frac{\frac{-dy}{dx}} {\frac{dy}{dx} (\frac{dy}{dx})}[/tex]
[tex]= \frac{1}{\frac{dy}{dx}}[/tex]
What happened to the minus sign?

Note that you can simplify 1/(dy/dx).
Nano-Passion said:
Also, would we be able to multiple out dy/dx if we had for example:

[tex]2 = \frac{1}{\frac{dy}{dx}}[/tex]
[tex]2(\frac{dy}{dx}) = 1[/tex]
Sure. And if the goal is to solve for dy/dx, divide both sides of the last equation by 2.
 
Mark44 said:
Sure.
What happened to the minus sign?
Simple mis-type.

Note that you can simplify 1/(dy/dx).
Like this?
[tex]=\frac{dx}{dy}[/tex]

Which would read change of x with respect to y.. Hmm, I'm not used to it being like that.

[tex]=\frac{dx}{dy} = lim_{Δx→0} \frac{Δx}{Δy}[/tex]
[tex]= lim_{Δx→0} \frac{Δx}{f(x+Δx) - f(x)}[/tex]
[tex]= f'(y)[/tex]

Is what I stated correct?

Sure. And if the goal is to solve for dy/dx, divide both sides of the last equation by 2.


Oh okay, thank you! I just wanted to make sure I wasn't violating any rules here pertaining to derivatives.
 
Last edited:
Nano-Passion said:
Simple mis-type.


Like this?
[tex]=\frac{dx}{dy}[/tex]

Which would read change of x with respect to y.. Hmm, I'm not used to it being like that.

[tex]=\frac{dx}{dy} = lim_{Δx→0} \frac{Δx}{Δy}[/tex]
No, the limit would be as Δy→0. Here the assumption would be that x is some function of y.
Nano-Passion said:
[tex]= lim_{Δx→0} \frac{Δx}{f(x+Δx) - f(x)}[/tex]
[tex]= f'(y)[/tex]

Is what I stated correct?




Oh okay, thank you! I just wanted to make sure I wasn't violating any rules here pertaining to derivatives.
 

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