Question

F[x,y[x]]

df= Df/dx dx/dx + Df/dy dy/dx

Answer 1

What's the question here?

Answer 2

what would be the second one?

Answer 3

I assume you're trying to write down the differential, dF

\[dF = \partial F/\partial x \ dx + \partial F/ \partial y \ dy \ ?\]

The thing is this is already just a function of F so the second partial derivative is zero.

Answer 4

...that is, assuming you have substituted all ys in the expression with x. I guess what I really need to know is what you want this expression for and then we can decide whether we should y substituted for or not.

Answer 5

Second last post: "The thing is this [the function F] is just a function of x", not F.

Answer 6

\[dF = F_x dx + F_y dy \]

\[d^2F =( F_{xx}+F_{xy}) dx + ( F_{xy}+F_{yy}) dy \]

Answer 7

Ah, no. That last expression you have written is definitely wrong.

The intuitive reason is the LHS is infinitesimals squared, but the RHS is just infinitesimals. The technical answer is differentials like dF, dx and dy live in what's called the de Rham Complex and the kind of expression you've written down is non-sensical.

Anyway -- what is it you are really trying to prove? We may not need to (or want to) go down this path.

Answer 8

F[x,y]=Cos[x y]

dF= -y Sin[x y] dx - x Sin[x y] dy

Answer 9

right ... so now what are you trying to calculate or characterize or show?

Answer 10

second order full differential

Answer 11

Ok. So if F = F(x,y). Then dF = F_x.dx + F_y.dy and then
d^2F = F_xx.(dx)^2 + 2F_xy.dx.dy + F_yy.(dy)^2

Notice this is true if and only if F_xy = F_yx, which it will be unless you're dealing with something very nasty.

Ok?

Answer 12

got it, thanks