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

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SUMMARY

The discussion centers around the differentiation of the function \(\phi(x,y)=0\) and the confusion surrounding the application of the chain rule. The correct interpretation is that taking the differential leads to the equation \(\phi_x \, dx + \phi_y \, dy = 0\), which is distinct from direct differentiation with respect to \(x\) or \(y\). The participants clarify that the initial equation presented by the user, \(\phi_x \, \frac{dx}{dy} + \phi_y \, \frac{dy}{dx} = 0\), incorrectly suggests simultaneous differentiation with respect to both variables. The proper approach involves recognizing the distinction between taking differentials and performing differentiation.

PREREQUISITES
  • Understanding of partial derivatives, specifically \(\phi_x\) and \(\phi_y\).
  • Familiarity with the chain rule in calculus.
  • Knowledge of differentials and their application in multivariable functions.
  • Basic grasp of implicit differentiation techniques.
NEXT STEPS
  • Study the concept of differentials in multivariable calculus.
  • Learn about implicit differentiation and its applications in solving equations.
  • Explore the chain rule in the context of multiple variables.
  • Review examples of taking differentials of functions to solidify understanding.
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Students and professionals in mathematics, particularly those studying calculus, multivariable functions, and differential equations. This discussion is beneficial for anyone looking to clarify the concepts of differentiation and differentials in the context of functions of multiple variables.

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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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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