Implicit Differentiation

This problem is inspired by this Wikipedia article.

Consider the graph of $f(x,y) = 0$ , where $f : \mathbb R^2 \to \mathbb R$ is a function that is continuously differentiable in a neighbourhood of the point $(x_0, y_0)$ , and $\displaystyle \frac{\partial f}{\partial y}(x_0, y_0) \neq 0$ .

Problem 1 (Implicit Function Theorem). Prove that there exists a unique differentiable function $\varphi$ such that $y_0 = \varphi(x_0)$ and $f(x, \varphi(x)) = 0$ in a neighbourhood of $x_0$ . In particular, denoting $y =\varphi(x)$ ,

$\displaystyle f_x + f_y \cdot \frac{\mathrm dy}{\mathrm dx} = 0.$

(Click for Solution)

Solution. Let $\varphi$ be any differentiable function such that $(t,\varphi(t))$ is in some neighborhood of $(x_0, y_0)$ without loss of generality. Define $g(t):= f(t,\varphi(t))$ . Then $g : \mathbb R \to \mathbb R$ is differentiable on a neighbourhood of $x_0$ . Using the chain rule,

$\begin{aligned} g'(t) &= f_x(t, \varphi(t)) \cdot \frac{\mathrm d}{\mathrm d t}(t) + f_y(t, \varphi(t) ) \cdot \frac{\mathrm d}{\mathrm d t}(\varphi(t)) \\ &= f_x(t, \varphi(t)) + f_y(t, \varphi(t) ) \cdot \varphi'(t). \end{aligned}$

On the other hand, by construction, $g = 0$ so that $g' = 0$ :

$\displaystyle f_x(t, \varphi(t)) + f_y(t, \varphi(t) ) \cdot \varphi'(t) = 0.$

Using algebruh, since $f_y \neq 0$ ,

$\displaystyle \varphi'(t) = -\frac{f_x(t, \varphi(t))}{f_y(t, \varphi(t) )} =: F(t, \varphi(t)).$

Since $F$ is continuous in the first argument and Lipschitz in the second, by the existence and uniqueness theorem in the theory of differential equations, there exists a unique differentiable $\varphi : \mathbb R \to \mathbb R$ such that $y_0 = \varphi(x_0)$ and $f(t, \varphi(t)) = 0$ in a neighbourhood of $x_0$ .