# 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 :)