ProofOfQuotientRule 2002-05-18 19:06:59 2004-04-24 20:04:55 Proof quotient rule 0 \usepackage{graphicx} %\usepackage{xypic} \usepackage{bbm} \newcommand{\Z}{\mathbbmss{Z}} \newcommand{\C}{\mathbbmss{C}} \newcommand{\R}{\mathbbmss{R}} \newcommand{\Q}{\mathbbmss{Q}} \newcommand{\mathbb}[1]{\mathbbmss{#1}} \newcommand{\figura}[1]{\begin{center}\includegraphics{#1}\end{center}} \newcommand{\figuraex}[2]{\begin{center}\includegraphics[#2]{#1}\end{center}} \newtheorem{dfn}{Definition} Let $F(x) = f(x)/g(x)$. Then \begin{eqnarray*} F'(x) & = & \lim_{h \to0} \frac{F(x+h)-F(x)}{h} = \lim_{h \to0} \frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}\\ & = & \lim_{h\to0} \frac{f(x+h)g(x)-f(x)g(x+h)}{hg(x+h)g(x)} \end{eqnarray*} Like the product rule, the key to this proof is subtracting and adding the same quantity. We separate $f$ and $g$ in the above expression by subtracting and adding the term $f(x)g(x)$ in the numerator. \begin{eqnarray*} F'(x) & = & \lim_{h \to0} \frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+h)}{hg(x+h)g(x)} \\ & = & \lim_{h \to0} \frac{g(x)\frac{f(x+h)-f(x)}{h}-f(x) \frac{g(x+h)-g(x)}{h}}{g(x+h)g(x)} \\ & = & \frac{\lim_{h \to0}g(x) \cdot \lim_{h \to0} \frac{f(x+h)-f(x)}{h}-\lim_{h \to0} f(x) \cdot \lim_{h \to0} \frac{g(x+h)-g(x)}{h}}{\lim_{h \to0}g(x+h) \cdot \lim_{h \to0}g(x)} \\ & = & \frac{g(x)f'(x)-f(x)g'(x)}{\lbrack g(x) \rbrack ^2} \end{eqnarray*}