Can Differentials Be Divided Out in Equations Involving Well-Behaved Functions?

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Homework Help Overview

The discussion revolves around the manipulation of differentials in the context of derivatives of well-behaved functions. The original poster presents an equality involving the derivatives of two functions, f and g, and questions whether it is valid to divide out the differentials to derive a relationship between the functions themselves.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the validity of dividing differentials in the context of derivatives, with some suggesting that this manipulation is not necessary for reaching conclusions about the functions. Others discuss the nature of differentials and their symbolic representation in calculus.

Discussion Status

The discussion is active, with participants providing insights into the treatment of differentials and derivatives. There is an exploration of whether formal calculations are required, and some participants suggest alternative approaches to arrive at similar conclusions without dividing differentials.

Contextual Notes

There is an assumption that the functions f and g are well-behaved and that the variables x and z are real numbers. The discussion also touches on the symbolic nature of differentials and the implications of treating derivatives as fractions.

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Homework Statement


Hi

Say I have the equality
[tex] \frac{df(x)}{dx} = \frac{dg(z)}{dz}[/tex]
where f and g are two functions that are well-behaved such that I can take their derivate. The variables x and z are both real, and run from -∞ to ∞. In this case, am I allowed to divide out the differentials dx and dz such that
[tex] df(x) = dg(z)[/tex]
which I can integrate and obtain
[tex] f(x) = g(z) + \text{constant}[/tex]
?Niles.
 
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Niles said:

Homework Statement


Hi

Say I have the equality
[tex] \frac{df(x)}{dx} = \frac{dg(z)}{dz}[/tex]
where f and g are two functions that are well-behaved such that I can take their derivate. The variables x and z are both real, and run from -∞ to ∞. In this case, am I allowed to divide out the differentials dx and dz such that
[tex] df(x) = dg(z)[/tex]
which I can integrate and obtain
[tex] f(x) = g(z) + \text{constant}[/tex]
?Niles.

I'm assuming from here on that f and g are functions of one variable...

You can come to this conclusion without doing any formal calculations (eg. "dividing out by differentials"). Although what you have done would work.

z is just a variable, so you can just write g(x) instead of g(z). Then, you get
[tex] \frac{df(x)}{dx}=\frac{dg(x)}{dx}.[/tex] Or, more sggestively, [itex]f'(x)=g'(x)[/itex]. Then, integrating both sides, you get [itex]f(x)=g(x)+constant[/itex].

Dividing out by differentials, which is a formal calculation (and thus not rigorous), is not necessary. Although, it is true that, in a sense, you can divide them out (see differential 1-forms).
 
Last edited:
Once we have the derivative, dy/dx, it is standard to define the "differential" by [tex]dy= (dy/dx)dx[/tex] where we can think of the "dx" as purely "symbolic". While the derivative is NOT a fraction, it is the limit of a fraction, the "difference quotient". It is possible to prove any "fraction property" by going back before the limit, applying the property to the difference quotient, then taking the limit. To make use of the fact that the derivative can be treated like a fraction, we define "differentials" in that way- we can say
[tex]f'(x)= \frac{dy}{dx}[/tex]
where the f'(x) on the left is the derivative and the dy/dx on the right is the ratio of the differentials.
 
Thanks for the quick replies.
 

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