Total differential for finding higer row derivatives

[irycio]

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.

 

[Mark44]

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?

 

[irycio]

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 :)

 

[Mark44]

I don't remember ever seeing anyone do what you're trying to do.

 

[irycio]

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]

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 :)