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

[bugatti79]

Hi Folks,

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

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

What am I doing wrong?

Thanks

 

[Ackbach]

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 $y$.

 

[bugatti79]

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

 

[Ackbach]

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?

 

[bugatti79]

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

 

[Ackbach]

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?