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This is perhaps something simple; but I am not quite getting why the implication is true; I seem to be missing something.

Supposedly, the implicit function theorem:

Let $f: \mathbb{R}^{n + m} \rightarrow \mathbb{R}^m$ be a continuously differentiable function, and let $\mathbb{R}^{n+m} $ have coordinates $( x, y)$, where $x \in \mathbb{R}^n$ and $y\in \mathbb{R}^m$. Fix a point $( a , b) = (a_1 , \ldots , a_n , b_1 , \ldots, b_m )$ with $ f( a, b) = c$, where $c \in \mathbb{R}^m$. If the matrix $( \partial f_i/\partial y_j)(a,b)$ is invertible, then there exists an open set $U$ containing $a$, and an open set $V$ contntaining $b$, and a unique continuously differentiable function $g: U \rightarrow V$ such that $$ \{ (\mathbf{x}, g(\mathbf{x}))|\mathbf x \in U \} = \{ (\mathbf{x}, \mathbf{y}) \in U \times V| f(\mathbf{x}, \mathbf{y}) = \mathbf{c} \}.$$

implies that implicit differentiation is okay:

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

What am I missing here? I again apologize if this is something very trivial.

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Let $ F: \mathbb{R}^{2} \to \mathbb{R} $ be a continuously differentiable function. Fix a point $ (a,b) \in \mathbb{R}^{2} $, and let $ c = F(a,b) $. Next, compute the Jacobian of $ F $: $$ \forall (x,y) \in \mathbb{R}^{2}: \quad [\mathbf{D}(F)](x,y) = \left[ \matrix{{\partial_{1} F}(x,y) & {\partial_{2} F}(x,y)} \right]. $$ If $ {\partial_{2} F}(a,b) \neq 0 $, then the Implicit Function Theorem implies that there exist

  • an open interval $ U $ containing $ a $,

  • an open interval $ V $ containing $ b $ and

  • a continuously differentiable function $ f: U \to V $

such that

  • $ f(a) = b $ and

  • $ \{ (x,y) \in U \times V ~|~ F(x,y) = c \} = \{ (x,f(x)) \in \mathbb{R}^{2} ~|~ x \in U \} $.

Now, define a function $ G: U \to \mathbb{R} $ by $$ \forall x \in U: \quad G(x) \stackrel{\text{def}}{=} F(x,f(x)). $$ Clearly, we have $$ \forall x \in U: \quad G(x) = c. $$ It thus follows from the Multivariable Chain Rule that \begin{align} \forall x \in U: \quad 0 &= G'(x) \\ &= {\partial_{1} F}(x,f(x)) \cdot 1 + {\partial_{2} F}(x,f(x)) \cdot f'(x) \\ &= 0. \end{align} As $ {\partial_{2} F}(a,b) \neq 0 $, we therefore obtain \begin{align} f'(a) &= - \frac{{\partial_{1} F}(a,f(a))}{{\partial_{2} F}(a,f(a))} \\ &= - \frac{{\partial_{1} F}(a,b)}{{\partial_{2} F}(a,b)}. \end{align}

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  • $\begingroup$ Ah, so the missing part was the chain rule. Thanks a lot. It is very clear now. $\endgroup$ – Rolanda Hooch Feb 22 '13 at 5:04
  • $\begingroup$ You're welcome! :) $\endgroup$ – Haskell Curry Feb 22 '13 at 5:05

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