Total differential for finding higer row derivatives

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Homework Help Overview

The discussion revolves around finding the second derivative \(\frac{d^2y}{dx^2}\) of an implicit function defined by the equation \(F: x^2 y^3=0\) using differential operators. Participants explore the implications of using total differentiation in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial differentiation leading to \(\frac{dy}{dx}=-\frac{2xy}{3x^2}\) and the subsequent application of the differential operator for the second time. Questions arise about extracting \(\frac{d^2y}{dx^2}\) from the resulting equation, with some participants suggesting implicit differentiation as an alternative method.

Discussion Status

There is ongoing exploration of different methods to find the second derivative, with some participants expressing familiarity with alternative techniques. One participant indicates they have resolved their question independently, while others continue to engage with the problem.

Contextual Notes

Participants note the complexity of applying total differentiation in systems of equations, highlighting potential challenges in finding higher derivatives in such contexts.

irycio
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Homework Statement


Well, let's take F: [tex]x^2 y^3=0[/tex].
Now, let's say thay y=y(x), y being an implicit function of x.
I want to find 2nd row derivative [tex]\frac{d^2y}{dx^2}[/tex]
using differential operator.


Homework Equations


not apply


The Attempt at a Solution


Using D for the first time:
[tex] 2xy^3dx+3x^2y^2dy=0[/tex]
Now I can find dy/dx:
[tex] \frac{dy}{dx}=-\frac{2xy}{3x^2}[/tex]

pretty simple, huh?

Now, using D for the 2nd time:

[tex] 2y^3dx^2+2xy^3d^2x+12xy^2dxdy+6x^2ydy^2+3x^2y^2d^2y=0[/tex]

Now, the question is: how to find the value of [tex]\frac{d^2y}{dx^2}[/tex] from the equation above. I know how to do it in another way, but I struggle to use that one.

Thanks in advance.
 
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irycio said:

Homework Statement


Well, let's take F: [tex]x^2 y^3=0[/tex].
Now, let's say thay y=y(x), y being an implicit function of x.
I want to find 2nd row derivative [tex]\frac{d^2y}{dx^2}[/tex]
using differential operator.


Homework Equations


not apply


The Attempt at a Solution


Using D for the first time:
[tex] 2xy^3dx+3x^2y^2dy=0[/tex]
Now I can find dy/dx:
[tex] \frac{dy}{dx}=-\frac{2xy}{3x^2}[/tex]

pretty simple, huh?

Now, using D for the 2nd time:

[tex] 2y^3dx^2+2xy^3d^2x+12xy^2dxdy+6x^2ydy^2+3x^2y^2d^2y=0[/tex]

Now, the question is: how to find the value of [tex]\frac{d^2y}{dx^2}[/tex] from the equation above. I know how to do it in another way, but I struggle to use that one.

Thanks in advance.

Why don't you start with this:
[tex] \frac{dy}{dx}=-\frac{2xy}{3x^2}[/tex]
and take the derivative implicitly with respect to x of both sides? Is this the other technique that you're struggling with?
 
I must have been misunderstood, apparently the lack of technical vocabulary can cause problems ;).

I'm familiar with the method you suggested, it's pretty simple and obvious. I was just wondering how to find the value of [tex]\frac{d^2y}{dx^2} [\tex] from the equation I got after differentiating the equation above totally for the 2nd time. I mean, it has to be possible somehow :)[/tex]
 
I don't remember ever seeing anyone do what you're trying to do.
 
Now, why I'm asking this. Of course it's as pointless as it may seem, but it seems to be quite a reasonable questions if a system of equations is to be considered.
Like...

u+v=x+y
x*sin(u)=y*sin(v)

where u=u(x,y), v=v(x,y). Now it's not that easy to find [tex]\frac{d^2u}{dx^2} or \frac{d^2v}{dy^2}[/tex] or whatever :). I mean, using total differentiation, one could easily handle this as a system of linear equations. At least I think so ;)
 
irycio said:

Homework Statement


Well, let's take F: [tex]x^2 y^3=0[/tex].
Now, let's say thay y=y(x), y being an implicit function of x.
I want to find 2nd row derivative [tex]\frac{d^2y}{dx^2}[/tex]
using differential operator.


Homework Equations


not apply


The Attempt at a Solution


Using D for the first time:
[tex] 2xy^3dx+3x^2y^2dy=0[/tex]
Now I can find dy/dx:
[tex] \frac{dy}{dx}=-\frac{2xy}{3x^2}[/tex]

pretty simple, huh?

Now, using D for the 2nd time:

[tex] 2y^3dx^2+2xy^3d^2x+12xy^2dxdy+6x^2ydy^2+3x^2y^2d^2y=0[/tex]

Now, the question is: how to find the value of [tex]\frac{d^2y}{dx^2}[/tex] from the equation above. I know how to do it in another way, but I struggle to use that one.

Thanks in advance.

Figured out the answer myself, that was a pretty tough puzzle, though :).
Lets's take our equation, keeping in mind, that y=y(x):
[tex] 2y^3dx^2+2xy^3d^2x+12xy^2dxdy+6x^2ydy^2+3x^2y^2d^2y=0[/tex]
Now, as x is a linear function of x, the 2nd differential of x disappears, hence:
[tex] 2xy^3d^2x=0[/tex]
now:
[tex] dy=\frac{dy}{dx} dx[/tex]
With that we can substitute all dy's (dy^2=dy*dy=(dy)^2 -obviously):
[tex] 2y^3dx^2+12xy^2 \frac{dy}{dx} dx^2 + 6x^2y (\frac{dy}{dx})^2 dx^2 + 3x^2y^2d^2y=0[/tex]
Then simplify and it's now easy to find
[tex] \frac{d^2y}{dx^2}[/tex]

Anyway, thanks for help, Mark44 :)
 

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