ProofOfChainRule 2002-05-31 16:25:35 2003-06-20 10:09:00 Proof chain rule 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let's say that $g$ is differentiable in $x_0$ and $f$ is differentiable in $y_0 = g(x_0)$. We define: \[ \varphi(y) = \left \{ \begin{array}{ll} \frac{f(y) - f(y_0)}{y-y_0} & \textrm{if $y \neq y_0$} \\ f'(y_0) & \textrm{if $y = y_0$} \end{array} \right. \] Since $f$ is differentiable in $y_0$, $\varphi$ is continuous. We observe that, for $x \neq x_0$, \[ \frac{f(g(x))-f(g(x_0))}{x-x_0} = \varphi(g(x)) \frac{g(x) - g(x_0)}{x-x_0}, \] in fact, if $g(x) \neq g(x_0)$, it follows at once from the definition of $\varphi$, while if $g(x) = g(x_0)$, both members of the equation are 0. Since $g$ is continuous in $x_0$, and $\varphi$ is continuous in $y_0$, \[ \lim_{x \to x_0} \varphi(g(x)) = \varphi(g(x_0)) = f'(g(x_0)), \] hence \begin{eqnarray*} (f \circ g)'(x_0) &=& \lim_{x\to x_0} \frac{f(g(x))-f(g(x_0))}{x-x_0} \\ &=& \lim_{x\to x_0} \varphi(g(x)) \frac{g(x) - g(x_0)}{x-x_0} \\ &=& f'(g(x_0))g'(x_0). \end{eqnarray*}