Implicit function theorem and derivative

Question

The implicit function theorem (in the $2$ dimensional case): Let $F\colon D \subset \Bbb R^2 \to \Bbb R^2$ and let $(x_0, y_0)$ be an interior point of $D$ with $F(x_0, y_0) = 0$. Suppose both first order partial derivatives of $F$ exist in $D$ and are continuous at $(x_0, y_0)$ with $F_y(x_0, y_0)= 0$. Then there is an interval $I \subset \Bbb R$, with $x_0$ an interior point of $I$, and a function $φ \colon I \to \Bbb R$, such that $φ$ is differentiable on $I$, $φ(x_0) = y_0$ and $F(x, φ(x)) = 0$ for each $x \in I$.

I don't see how I can get from the chain rule that $$φ'(x)=\frac{-F_x(x,φ(x))}{F_y(x,φ(x))}.$$ I know that it should follow from the chain rule, but why?

(I stated here the theorem for the $2$ dimensional case, but of course we look at it in the $n$-dimensional case).

Answer 1

Given that for all $x\in I$ we have $F(x,\varphi(x))=0$, we differentiate with respect to $x$: $$0 = \frac{d}{dx}F(x,\varphi(x))=F_{x}(x,\varphi(x)) + F_{y}(x,\varphi(x))\varphi'(x).$$ Solving for $\varphi'(x)$ gives us the desired result: $$\varphi'(x) = \frac{-F_{x}(x,\varphi(x))}{F_{y}(x,\varphi(x))}.$$

Answer 2

By chain rule: $$D (F\circ(\operatorname{Id},\varphi))(x)=DF(x,\varphi(x))\cdot\begin{pmatrix}1 \\ \varphi'(x)\end{pmatrix}=D_xF(x,\varphi(x))+D_y F(x,\varphi(x))\cdot\varphi'(x)$$ Also, $D(F\circ(\operatorname{Id},\varphi))(x)=0$ as $F(x,\varphi(x))=0$ for all $x\in I$, so $F\circ(\operatorname{Id},\varphi)$ is constant. Then you get the result.