Relation between chain rule and implicit differentiation derivation in multi variable calculus

Question

So my question is on the derivation of the implicit differentiation (taken from here).

The general chain rule, from here, it says that if we have a function $z$ of $n$ variables, $x_1, x_2,\ldots,x_n$ and each of these variables are in turn a function of $m$ variables, $t_1, t_2,\ldots, t_m$. Then for any $t_i, i=1, 2, \ldots, m$ we have (1) $$ \frac{\partial z}{\partial t_i}=\frac{\partial z}{\partial x_1}\frac{\partial x_1}{\partial t_i}+\frac{\partial z}{\partial x_2}\frac{\partial x_2}{\partial t_i}+\cdots+\frac{\partial z}{\partial x_n}\frac{\partial x_n}{\partial t_i}$$

My question is how does did the differentiation of $F$ with respect to $x$ come up as it did. How did it come up as (2) $$\frac{\partial F}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial F}{\partial y}\frac{\partial y}{\partial x}+\cdots+\frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=0$$ If I match (2) with (1), then it seems that the left hand side is $\frac{\partial F}{\partial x}$ because $F=z, x=x_1, y=x_2, z=x_3$. As far as the initial condition of $F$ having three variables $(x, y, z)$ that in turn are each supposed to be functions of more variables, can it be assumed that $x$ and $y$ are constants; this would be mean effectively that $x=g(t_1, t_2, t_3)=x$ of some function g and $y=y(t_1, t_2, t_3)=y$. So is the left hand side of (2) $\frac{\partial F}{\partial x}$ or some different notation. What does the differentiation of $F$ with respect to $x$ on the left hand side look like in symbols then?

Answer 1

When god created the world there were three independent quantities $x$, $y$, $z$ changing with time, say. Then a physicist came and said: These three quantities are related by a so-and-so equation $$F(x,y,z)=0\ .\tag{1}$$ At any rate, at some instant $t_0$ of time one indeed had $F(x_0,y_0,z_0)=0$. The equation $(1)$ connecting the three quantities $x$, $y$, $z$ reduces the degrees of freedom by $1$. Therefore we have the feeling that, given two values $x$ and $y$, the value of $z$ should be determined by $(1)$, at least in the neighborhood of the point $P_0=(x_0,y_0,z_0)$. This means that there should be a function $z=\psi(x,y)$ defined in a neighborhood $U$ of $(x_0,y_0)$ and giving $\psi(x_0,y_0)=z_0$, such that $$F\bigl(x,y, \psi(x,y)\bigr)=0\qquad\forall (x,y)\in U\ .\tag{2}$$ Sometimes we can explicitly solve $(1)$ for $z$, and then we have this $\psi$ as a a function of the variables $x$ and $y$ "in finite terms". But in most cases we can't. That's where the implicit function theorem comes in. It says that such a function $z=\psi(x,y)$ defined in the neighborhood of $(x_0,y_0)$ indeed exists, and that this $\psi$ is even differentiable. In particular, we are able to compute the partial derivatives $\psi_x(x_0,y_0)$ and $\psi_y(x_0,y_0)$ from the given data. The chain rule allows to argue as follows: From $(2)$ we obtain by differentiating with respect to $x$ and putting $(x,y):=(x_0,y_0)$ afterwards: $$F_{.1}(x_0,y_0,z_0)\cdot 1+F_{.3}(x_0,y_0,z_0)\psi_x(x_0,y_0)=0\ .$$ This gives $$\psi_x(x_0,y_0)=-{F_{.1}(x_0,y_0,z_0)\over F_{.3}(x_0,y_0,z_0)}\ ,\tag{3}$$ under the "technical assumption" $$F_{.3}(x_0,y_0,z_0)\ne0\ .\tag{4}$$ The formula $(3)$ is usually written in the shorthand version $${\partial z\over\partial x}\biggr|_{(x_0,y_0)}=-{F_x\over F_z}\biggr|_{(x_0,y_0,z_0)}\ .$$ Similarly, one has $${\partial z\over\partial y}\biggr|_{(x_0,y_0)}=-{F_y\over F_z}\biggr|_{(x_0,y_0,z_0)}\ .$$ Note that the assumption $(4)$ is essential for all this to hold.

Answer 2

Actually it will be much easier if you use total differential concept $$F\big(x,y,z\big)=0$$ $$dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}dz=0$$ Note that $z=z(x,y)$ and therefore $$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$ If $dz$ plugged in $$dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy+\frac{\partial F}{\partial z}\bigg(\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy\bigg)=0$$ $$dF=\bigg(\frac{\partial F}{\partial x}+\frac{\partial F}{\partial z} \frac{\partial z}{\partial x}\bigg)dx+\bigg(\frac{\partial F}{\partial y}+\frac{\partial F}{\partial z}\frac{\partial z}{\partial y} \bigg)dy=0$$ The equality holds for any $dx$ and $dy$ if $$\bigg(\frac{\partial F}{\partial x}+\frac{\partial F}{\partial z} \frac{\partial z}{\partial x}\bigg)=0 \Rightarrow\frac{\partial z}{\partial x}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}$$ $$\bigg(\frac{\partial F}{\partial y}+\frac{\partial F}{\partial z} \frac{\partial z}{\partial y}\bigg)=0 \Rightarrow\frac{\partial z}{\partial y}=-\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}$$