# Implicit function theorem and implicit differentiation

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