How do you obtain this form of the second derivative of this multivariable function?

Question

I'm studying a book of original differential equations, and the excerpt below is given as the general form of high order derivatives for a multivariable function.

I have two questions:

1) Why isn't there a full differential beside the ${\frac{\delta F}{\delta x}}$ term? Shouldn't there be a $dx$ there?

2) Every time I do the second derivative, using the chain rule, it looks nothing like the second line. How is that derivative formed?

Answer 1

HINT :

$$F(x,y,C_1,..., C_n)=0 \tag 1$$ FIRST EQUATION :

Total differential of Eq.$(1)$ : $$dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy=0$$ So, you obtain the first equation : $$\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}\frac{dy}{dx}=0 \tag 2$$

SECOND EQUATION :

Let $\quad \Phi(x,y)=\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}\frac{dy}{dx}=0$

The relationship $(2)$ is valid for any function $F(x,y)=0$. So, it is valid for the function $\Phi(x,y)=0$ as well : $$\frac{\partial \Phi}{\partial x}+\frac{\partial \Phi}{\partial y}\frac{dy}{dx}=0$$ We have to compute $\frac{\partial \Phi}{\partial x}$ and $\frac{\partial \Phi}{\partial y}$. Using the chain rule :

$\frac{\partial \Phi}{\partial x}=\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}\frac{dy}{dx} \right) = \frac{\partial^2 F}{\partial x^2}+\frac{\partial^2 F}{\partial y\partial x}\frac{dy}{dx}+\frac{\partial F}{\partial y}\frac{d^2y}{dx^2}$

$\frac{\partial \Phi}{\partial y}=\frac{\partial}{\partial y}\left(\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}\frac{dy}{dx} \right) = \frac{\partial^2 F}{\partial x\partial y}+\frac{\partial^2 F}{\partial y^2}\frac{dy}{dx}$

$$\frac{\partial \Phi}{\partial x}+\frac{\partial \Phi}{\partial y}\frac{dy}{dx}=0=\frac{\partial^2 F}{\partial x^2}+\frac{\partial^2 F}{\partial y\partial x}\frac{dy}{dx}+\frac{\partial F}{\partial y}\frac{d^2y}{dx^2} + \left(\frac{\partial^2 F}{\partial x\partial y}+\frac{\partial^2 F}{\partial y^2}\frac{dy}{dx}\right)\frac{dy}{dx}$$ After simplification you obtain the second equation : $$\frac{\partial^2 F}{\partial x^2}+2\frac{\partial^2 F}{\partial y\partial x}\frac{dy}{dx} + \frac{\partial F}{\partial y}\frac{d^2y}{dx^2} +\frac{\partial^2 F}{\partial y^2}\left(\frac{dy}{dx}\right)^2=0$$

To continue on the same way, let $\Phi(x,y)=\frac{\partial^2 F}{\partial x^2}+2\frac{\partial^2 F}{\partial y\partial x}\frac{dy}{dx} + \frac{\partial F}{\partial y}\frac{d^2y}{dx^2} +\frac{\partial^2 F}{\partial y^2}\left(\frac{dy}{dx}\right)^2=0$ and differentiate $\frac{\partial \Phi}{\partial x}+\frac{\partial \Phi}{\partial y}\frac{dy}{dx}=0$ as above. And so on, the calculus is similar at each step.

Answer 2

We differentiate the composite function $F(x, y(x), C_1, \dots, C_2)$ (of one variable $x$). The dependence of that function on $x$ is twofold: first, "directly" on $x$, second, on $y$, which in its turn depends on $x$. Hence the first formula.

Now, let's write the variables: $\frac{\partial F}{\partial x}(x, y(x)) + \frac{\partial F}{\partial y}(x, y(x)) \frac{dy}{dx}(x)$. Differentiating it in $x$ we obtain $\frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y \partial x} \frac{dy}{dx} + \frac{\partial^2 F}{\partial x \partial y} \frac{dy}{dx} + \frac{\partial^2 F}{\partial x^2} \frac{dy}{dx} \frac{dy}{dx} + \frac{\partial F}{\partial y} \frac{d^2 y}{dx^2}$, and use symmetry of mixed derivatives to get the second formula.