Homework Help: Total differential for finding higer row derivatives

1. Apr 22, 2010

irycio

1. The problem statement, all variables and given/known data
Well, let's take F: $$x^2 y^3=0$$.
Now, let's say thay y=y(x), y being an implicit function of x.
I want to find 2nd row derivative $$\frac{d^2y}{dx^2}$$
using differential operator.

2. Relevant equations
not apply

3. The attempt at a solution
Using D for the first time:
$$2xy^3dx+3x^2y^2dy=0$$
Now I can find dy/dx:
$$\frac{dy}{dx}=-\frac{2xy}{3x^2}$$

pretty simple, huh?

Now, using D for the 2nd time:

$$2y^3dx^2+2xy^3d^2x+12xy^2dxdy+6x^2ydy^2+3x^2y^2d^2y=0$$

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

2. Apr 22, 2010

Staff: Mentor

$$\frac{dy}{dx}=-\frac{2xy}{3x^2}$$
and take the derivative implicitly with respect to x of both sides? Is this the other technique that you're struggling with?

3. Apr 22, 2010

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 $$\frac{d^2y}{dx^2} [\tex] from the equation I got after differentiating the equation above totally for the 2nd time. I mean, it has to be possible somehow :) 4. Apr 22, 2010 Mark44 Staff: Mentor I don't remember ever seeing anyone do what you're trying to do. 5. Apr 22, 2010 irycio Now, why I'm asking this. Of course it's as pointless as it may seem, but it seems to be quite a reasonable questions if a system of equations is to be considered. Like... u+v=x+y x*sin(u)=y*sin(v) where u=u(x,y), v=v(x,y). Now it's not that easy to find [tex] \frac{d^2u}{dx^2} or \frac{d^2v}{dy^2}$$ or whatever :). I mean, using total differentiation, one could easily handle this as a system of linear equations. At least I think so ;)

6. Apr 22, 2010

irycio

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

Anyway, thanks for help, Mark44 :)