Is this derivation of implicit differentiation using chain rule is technically valid?

Question

In my text book I found a derivation of implicit differentiation using the chain rule to get this formula:

$$\frac{dy}{dx}=\frac{-\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$$

Which: $x=x,y=g(x),F=F(x,y)$

I tried to derive same formula as following:

$$\frac{dy}{dx}=\frac{\partial y}{\partial F}\frac{\partial F}{\partial x}+\frac{\partial y}{\partial y}\frac{\partial y}{\partial x}+\frac{\partial y}{\partial x}\frac{\partial x}{\partial x}$$ $$\frac{dy}{dx}=\frac{\partial y}{\partial F}\frac{\partial F}{\partial x}+\frac{\partial y}{\partial x}+\frac{\partial y}{\partial x}$$ $$\frac{dy}{dx}=\frac{\partial y}{\partial F}\frac{\partial F}{\partial x}+2\frac{\partial y}{\partial x}$$

then substracting $\frac{\partial y}{\partial x}$ from each side: $$0=\frac{\partial y}{\partial F}\frac{\partial F}{\partial x}+\frac{\partial y}{\partial x}$$ $$\frac{\partial y}{\partial x}=-\frac{\partial y}{\partial F}\frac{\partial F}{\partial x}$$ $$\frac{\partial y}{\partial x}=\frac{-\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$$

I think something is technically wrong in the step of subtracting but I'm not sure. Is my derivation valid ?

Thanks

Answer 1

Look at your first line, in the case where $g(x)-3x, F(x,y) = x^2y^3$. It does not appear to be true.

Answer 2

Your notation is terribly confusing, so let me do things properly for you.

You have a point $(a,b)$ such that $F(a,b) = 0$. You also have that $\frac {\partial F} {\partial y} (a,b) \ne 0$. The implicit function theorem guarantees the existence of a differentiable function $g$ such that $y = g(x)$ in some neighbourhood of $a$ and $F(x,g(x)) = 0$ on it. You want to find $g'$ on this neighbourhood.

Begin by writing that $0 = F(x,g(x))$, and derive with respect to $x$:

$$0 = (0)' = F(x,g(x))' = \frac {\partial F} {\partial x} + \frac {\partial F} {\partial y} g'(x) ,$$

whence it follows that

$$g'(x) = - \frac {\frac {\partial F} {\partial x}} {\frac {\partial F} {\partial y}} .$$

Answer 3

The interchanging of $\partial x, \partial y$ and $\partial F$ cannot be done without making additional presumptions. You can achieve the result by using the total differentiation.

$dF=\frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy=0$

The equation has to be set equal to zero, because this is the condition for a stationary point $(x_0,y_0)$.

$-\frac{\partial F}{\partial x}dx = \frac{\partial F}{\partial y}dy$

Dividing the equation by $\frac{\partial F}{\partial y}$ and $dx$

$\Large{-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}}=\normalsize{\frac{dy}{dx}}$

This equation has to be true for a stationary point. $F(x,y)$ has to be differentiable in $(x_0,y_0)$.