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I have the following question concerning the implicit function theorem (and, I think, more generally the behavior of a differential operator, such as the one used for derivatives).

[I hope the question does not sound too trivial, but being self-thaught I am often puzzled by what are usually trivial issues]

Implicit Function
We start from an arbitrary implicit function $$F(x,y)=0$$ We assume that $y$ can be expressed as $y=f(x)$, hence we can write $$F(x, f(x))=0.$$ Now, if we want to have $\frac{dF}{dx}$ we have $$\require{cancel}\frac{dF}{dx} \equiv F_x \frac{\cancel{dx}}{\cancel{dx}} + F_y \frac{dy}{dx} = 0 \tag{*}$$ which leads to $$ \frac{dy}{dx}=-\frac{F_x}{F_y}.$$

Fair enough. The question is the following:

Why do we have the $0$ on the RHS of (*)?
[what I mean is that when we focus on the total derivative of a standard function $F(x,y)$, there is no $0$ anywhere]
Does it come from the fact that the differential operator $\frac{d}{dx}$ acts basically in the following way: $$\frac{d}{dx}F(x,f(x)) = \frac{d}{dx}0 \implies F_x + F_y \frac{dy}{dx} = 0 ?$$

Any feedback is most welcome.
Thank you for your time.

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  • $\begingroup$ the cancelation above is wrong... what happen is that $\frac{dx}{dx}=1$ $\endgroup$ – janmarqz Nov 22 '15 at 16:58
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    $\begingroup$ @janmarqz: Thanks a lot. Actually, this is what I meant. Beyond that point, are the remaining things correct? (in particular the very last formula). $\endgroup$ – Kolmin Nov 22 '15 at 17:00
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    $\begingroup$ sure, it is good $\endgroup$ – janmarqz Nov 22 '15 at 17:02
  • $\begingroup$ @janmarqz Thanks a lot for the feedback. $\endgroup$ – Kolmin Nov 22 '15 at 17:03
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You can use total derivative as below

$$dF(x,y=f(x))=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}\frac{dy}{dx}dx=0$$ $$\Rightarrow \frac{dy}{dx}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$$

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