\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.4in,right=0.5in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue} \Large Math 207} \lfoot{\small \copyright $\;$ copyright Hidegkuti, Powell, 2012} \cfoot{} \chead{\LARGE Proving the Product Rule} \rfoot{\small Last revised: October 24, 2013} \rhead{\large page \thepage } \textwidth 7.5in \textheight 9.6in \setlength{\headheight}{27pt} \setlength{\parindent}{0pt} \begin{document} %TCIMACRO{% %\TeXButton{box start - 7.5in}{\begin{boxedminipage}{7.5in} %\setlength{\fboxrule}{12pt} %\setlength{\fboxsep}{15pt} %\large}}% %BeginExpansion \begin{boxedminipage}{7.5in} \setlength{\fboxrule}{12pt} \setlength{\fboxsep}{15pt} \large% %EndExpansion \textbf{Theorem 1:} \ If $f$ is differentiable at $a$, then it is continuous there.% %TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion \medskip Proof: Suppose that $f$ is differentiable at a number $a$. \ Then $f^{\prime }\left( a\right) $ exists which means that $f\left( a\right) $ exists and the limit $\lim\limits_{h\rightarrow 0}\dfrac{f\left( a+h\right) -f\left( a\right) }{h}$ also exists and is finite. \ Let us start with the true statement that $0=0\cdot f^{\prime }\left( a\right) $.% \begin{eqnarray*} 0 &=&0\cdot f^{\prime }\left( a\right) \\ 0 &=&\lim\limits_{h\rightarrow 0}h\cdot \lim\limits_{h\rightarrow 0}\dfrac{% f\left( a+h\right) -f\left( a\right) }{h}\text{ \ \ \ \ \ \ \ \ \ \ \ by the product rule of limits} \\ 0 &=&\lim\limits_{h\rightarrow 0}\left( h\cdot \dfrac{f\left( a+h\right) -f\left( a\right) }{h}\right) \text{ \ \ \ \ \ \ \ \ \ \ \ cancel out }h \\ 0 &=&\lim\limits_{h\rightarrow 0}\left( f\left( a+h\right) -f\left( a\right) \right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by the difference rule of limits} \\ 0 &=&\lim\limits_{h\rightarrow 0}f\left( a+h\right) -\lim\limits_{h\rightarrow 0}f\left( a\right) \\ \lim\limits_{h\rightarrow 0}f\left( a\right) &=&\lim\limits_{h\rightarrow 0}f\left( a+h\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by the constant rule of limits} \\ f\left( a\right) &=&\lim\limits_{h\rightarrow 0}f\left( a+h\right) \end{eqnarray*}% and $f\left( a\right) =\lim\limits_{h\rightarrow 0}f\left( a+h\right) $ means that $f$ is continuous at $a$.\medskip %TCIMACRO{% %\TeXButton{box start - 7.5in}{\begin{boxedminipage}{7.5in} %\setlength{\fboxrule}{12pt} %\setlength{\fboxsep}{15pt} %\large}}% %BeginExpansion \begin{boxedminipage}{7.5in} \setlength{\fboxrule}{12pt} \setlength{\fboxsep}{15pt} \large% %EndExpansion \textbf{Theorem 2:} \ (Product rule of derivatives) \ If $f$ and $g$ are differentiable at $x$, then so is $fg$ and \qquad \qquad \qquad \qquad the derivative is $\left( fg\right) ^{\prime }=f^{\prime }g+fg^{\prime }$.% %TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion \medskip Proof: \ Suppose that $f$ and $g$ are differentiable at $x$. \ \begin{eqnarray*} \left( fg\right) ^{\prime } &=&\lim\limits_{h\rightarrow 0}\dfrac{fg\left( x+h\right) -fg\left( x\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{% f\left( x+h\right) g\left( x+h\right) -f\left( x\right) g\left( x\right) }{h}% \text{ \ \ \ \ \ \ \ \ \ \ \ \ smuggle in }f\left( x+h\right) g\left( x\right) \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right) g\left( x+h\right) -f\left( x+h\right) g\left( x\right) +f\left( x+h\right) g\left( x\right) -f\left( x\right) g\left( x\right) }{h} \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right) \left[ g\left( x+h\right) -g\left( x\right) \right] +g\left( x\right) \left[ f\left( x+h\right) -f\left( x\right) \right] }{h} \\ &=&\lim\limits_{h\rightarrow 0}f\left( x+h\right) \dfrac{g\left( x+h\right) -g\left( x\right) }{h}+g\left( x\right) \dfrac{f\left( x+h\right) -f\left( x\right) }{h}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by the sum rule of limits} \\ &=&\lim\limits_{h\rightarrow 0}f\left( x+h\right) \dfrac{g\left( x+h\right) -g\left( x\right) }{h}+\lim\limits_{h\rightarrow 0}g\left( x\right) \dfrac{% f\left( x+h\right) -f\left( x\right) }{h}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ by the product rule of limits} \\ &=&\lim\limits_{h\rightarrow 0}f\left( x+h\right) \cdot \lim\limits_{h\rightarrow 0}\dfrac{g\left( x+h\right) -g\left( x\right) }{h}% +\lim\limits_{h\rightarrow 0}g\left( x\right) \lim\limits_{h\rightarrow 0}% \dfrac{f\left( x+h\right) -f\left( x\right) }{h}\text{\ \ \ \ by the constant rule of limits} \\ &=&\lim\limits_{h\rightarrow 0}f\left( x+h\right) \cdot \lim\limits_{h\rightarrow 0}\dfrac{g\left( x+h\right) -g\left( x\right) }{h}% +g\left( x\right) \lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right) -f\left( x\right) }{h} \end{eqnarray*}% We now realize the following:% \begin{equation*} \lim\limits_{h\rightarrow 0}\dfrac{g\left( x+h\right) -g\left( x\right) }{h}% =g^{\prime }\left( x\right) \text{ \ \ \ and \ \ }\lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right) -f\left( x\right) }{h}=f^{\prime }\left( x\right) \end{equation*}% Furthermore, $f$ is continuous because it is differentiable, and so $% \lim\limits_{h\rightarrow 0}f\left( x+h\right) =f\left( x\right) $. \ Thus we now have that% \begin{eqnarray*} \left( fg\right) ^{\prime } &=&\lim\limits_{h\rightarrow 0}f\left( x+h\right) \cdot \lim\limits_{h\rightarrow 0}\dfrac{g\left( x+h\right) -g\left( x\right) }{h}+g\left( x\right) \lim\limits_{h\rightarrow 0}\dfrac{% f\left( x+h\right) -f\left( x\right) }{h} \\ &=&~~~~~~f\left( x\right) ~~~~~\cdot ~~~~~~g^{\prime }\left( x\right) ~~~~~~~~~~~~+g\left( x\right) \cdot ~~~~~~~~~f^{\prime }\left( x\right) \end{eqnarray*} {\small %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}