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I won't post the whole rigorous statement of the theorem, but basically the theorem states that

If ##F(x,y) = 0## on a neighborhood of the form ##[x-\delta ,x+\delta ]\times [y- \epsilon ,y+\epsilon ]## and if ##\frac{\partial F(x,y)}{\partial y} \neq 0##, then there exists a function ##y=\phi (x)\: s.t. F(x,\phi (x)) = 0 \: \forall x \in [x-\delta ,x+\delta ]##. My question is about ##\phi '(x)##, which can be obtained by deriving ##F(x,\phi (x))## w.r.t. x as follows:

[tex]

\frac{d}{dx}F(x,\phi (x)) = F_x(x,\phi (x)) + F_y(x,\phi (x))\phi '(x) = 0\\\rightarrow \phi '(x) = -\frac{F_x(x,\phi (x))}{F_y(x,\phi (x))}

[/tex]

Why can we set ##\frac{d}{dx}F(x,\phi (x)) = 0##? By definition, by composing F with ##\phi##, we are on a curve where F is 0 for all x(and so it makes sense that all its derivatives w.r.t. x are 0). But ##\frac{\partial F(x,y)}{\partial x}## isn't necessarily equal to 0 in that same neighborhood. What is the difference between the two? Isn't ##F(x,\phi (x)) \in F(x,y)##? I'm quite confused

If ##F(x,y) = 0## on a neighborhood of the form ##[x-\delta ,x+\delta ]\times [y- \epsilon ,y+\epsilon ]## and if ##\frac{\partial F(x,y)}{\partial y} \neq 0##, then there exists a function ##y=\phi (x)\: s.t. F(x,\phi (x)) = 0 \: \forall x \in [x-\delta ,x+\delta ]##. My question is about ##\phi '(x)##, which can be obtained by deriving ##F(x,\phi (x))## w.r.t. x as follows:

[tex]

\frac{d}{dx}F(x,\phi (x)) = F_x(x,\phi (x)) + F_y(x,\phi (x))\phi '(x) = 0\\\rightarrow \phi '(x) = -\frac{F_x(x,\phi (x))}{F_y(x,\phi (x))}

[/tex]

Why can we set ##\frac{d}{dx}F(x,\phi (x)) = 0##? By definition, by composing F with ##\phi##, we are on a curve where F is 0 for all x(and so it makes sense that all its derivatives w.r.t. x are 0). But ##\frac{\partial F(x,y)}{\partial x}## isn't necessarily equal to 0 in that same neighborhood. What is the difference between the two? Isn't ##F(x,\phi (x)) \in F(x,y)##? I'm quite confused

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