Yes, that makes sense! Thank you for explaining it to me.

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Discussion Overview

The discussion centers around the differentiation of the function \(\phi(x,y)=0\) and the interpretation of its differential form. Participants explore the implications of applying the chain rule and the concept of taking differentials versus differentiation in the context of multivariable calculus.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents an equation derived from the chain rule, \(\phi_x \, \frac{dx}{dy} + \phi_y \, \frac{dy}{dx} = 0\), and questions its validity.
  • Another participant challenges the clarity of the first participant's equation, suggesting it appears to mix differentiation with respect to \(x\) and \(y\) simultaneously.
  • Some participants clarify that \(\phi_x \, dx + \phi_y \, dy = 0\) represents taking the differential of both sides, and provide alternative forms of the equation when differentiating with respect to \(x\) or \(y\).
  • There is a discussion about the distinction between taking differentials and differentiation, with one participant emphasizing that the former is not strictly the same as differentiation.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the equations and the process of differentiation versus taking differentials. No consensus is reached regarding the initial equation presented by the first participant.

Contextual Notes

Some participants note that the terminology used may lead to confusion, particularly regarding the distinction between taking differentials and performing differentiation. The discussion also highlights the need for clarity in the application of the chain rule in multivariable contexts.

bugatti79
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Hi Folks,

It is been given that differentiation of \phi(x,y)=0 is \phi_{x} dx+ \phi_{y} dy=0 however I arrive at

\phi_{x} dx/dy+ \phi_{y} dy/dx=0 via the chain rule. Where \phi_{x}=d \phi/dx etc

What am I doing wrong?

Thanks
 
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What are you differentiating $\phi$ with respect to? Your equation
$$\phi_x \, \frac{dx}{dy}+\phi_y \, \frac{dy}{dx}=0$$
looks schizophrenic - as if you're simultaneously trying to differentiate w.r.t. $x$ and $y$.
 
Ackbach said:
What are you differentiating $\phi$ with respect to? Your equation
$$\phi_x \, \frac{dx}{dy}+\phi_y \, \frac{dy}{dx}=0$$
looks schizophrenic - as if you're simultaneously trying to differentiate w.r.t. $x$ and $y$.

Hi, I am referring to eqn 3.2.9 in attached
 

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    Implicit 1.jpg
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I would say that $\phi_x \, dx+\phi_y \, dy=0$ is really taking the differential of both sides. If you wanted to differentiate w.r.t. $x$, then you would get
$$\phi_x+\phi_y \, \frac{dy}{dx}=0,$$
and if w.r.t. $y$, you'd get
$$\phi_x \, \frac{dx}{dy}+\phi_y=0.$$

Does that answer your question?
 
Ackbach said:
I would say that $\phi_x \, dx+\phi_y \, dy=0$ is really taking the differential of both sides. If you wanted to differentiate w.r.t. $x$, then you would get
$$\phi_x+\phi_y \, \frac{dy}{dx}=0,$$
and if w.r.t. $y$, you'd get
$$\phi_x \, \frac{dx}{dy}+\phi_y=0.$$

Does that answer your question?

I think the last 2 equations are coming from the chain rule so yes I can see that.

However, not sure how one arrives at the first equation 3.2.9 by differentiating "both sides"?

Thanks
 
bugatti79 said:
I think the last 2 equations are coming from the chain rule so yes I can see that.

However, not sure how one arrives at the first equation 3.2.9 by differentiating "both sides"?

Thanks

It was slightly sloppy wording, in my view. They're really "taking the differential" of both sides. Imagine that $y=f(x)$. If you "took the differential" of both sides, you'd get $dy=f'(x) \, dx$, right? This is not, strictly speaking, differentiation. It's taking the differential. If you were to differentiate, you'd get $y'=f'(x)$.

So, taking the differential of $\phi(x,y)=0$ amounts to doing $\phi_x \, dx+ \phi_y \, dy=0$ in a similar fashion. Does that make sense?
 
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