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math_handout/Math_Handout_CHEG231.tex

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+\documentclass[11pt]{article}
+
+\usepackage[margin=1in]{geometry}
+
+\usepackage[bookmarks=true]{hyperref}
+\hypersetup{
+	pdftitle={},
+	pdfauthor={},
+	pdfkeywords={},
+	bookmarksnumbered,
+	breaklinks=true,
+	urlcolor=blue,
+	citecolor=black,
+	colorlinks=true,
+	linkcolor=black,
+}
+
+\usepackage{amsmath}
+\usepackage{amssymb}
+
+\usepackage{booktabs}
+\usepackage{caption}
+\captionsetup[table]{skip=1ex}
+
+\usepackage{graphicx}
+%\graphicspath{{figures/}}
+
+\usepackage{parskip}
+%\usepackage{indentfirst}
+%\setlength{\parindent}{2em}
+
+\pagestyle{plain}
+
+\begin{document}
+
+% Header text
+\hspace{-0.025\textwidth}\parbox[b]{1.05\textwidth}{\centering \bfseries
+\vspace{-2\baselineskip}  % I move it up a little bit above the 1" margins
+{\Large \uppercase{CHEG 231 Chemical Engineering Thermodynamics}}\\
+{\large {Department of Chemical and Biomolecular Engineering}}\\
+{\large {University of Delaware}}
+}
+
+\vskip 2\baselineskip
+
+\begin{center}
+
+{\uppercase{\Large Math reminders}\par}
+
+%{\itshape Eric\ M.\ Furst\par}
+{\itshape \today \par}
+
+\vskip \baselineskip
+
+%{\large Nome: \hrulefill\hspace{1em}{\small RA}: \rule{25mm}{0.4pt}}
+
+\vskip \baselineskip
+
+%\begin{minipage}{10cm}
+%\itshape\small\emph{Instruções:} Resolva todas as questões.
+%Mostre todo o seu trabalho e considerações feitas.
+%Inclua unidades apropriadas para todas as respostas e direções para grandezas vetoriais.
+%Integridade acadêmica é esperada de todos.
+%\end{minipage}
+
+\end{center}
+
+%Grupo (nome e {\small RA}):\\\answerbox{\linewidth}{25mm}
+
+%f(x) = \frac{1}{x}
+\section{A few common integrals} 
+Basic integration and integrals involving functions like $\frac{1}{x}$ show up frequently in thermodynamics. 
+\begin{center}
+\renewcommand{\arraystretch}{1.5}
+\setlength{\tabcolsep}{1em}
+\begin{tabular}{cc}\\
+%\multicolumn{4}{c}{Fourier transforms} \\
+$f(x)$ & $F(x) = \int f(x) dx$ \\
+\hline
+$x$ & $\frac{1}{2}x^2$ \\
+$\frac{1}{x}$ & $\ln x$ \\
+$e^x$ & $e^x$ \\
+%\multicolumn{4}{l}{\footnotesize $^1${$H(t)$ is the Heaviside step function.}} \\
+%\multicolumn{4}{l}{\footnotesize $^2${$\Gamma(x) = \int^\infty_0 e^r r^{x-1}dr$ is the Gamma function. If $x$ is a positive integer, then $\Gamma(x+1) = x!$}} \\
+\end{tabular}
+\end{center}
+
+\vspace{2\baselineskip}
+Remember that the indefinite integral of a function includes a constant term, $\int f(x) dx = F(x) + c$ were $f(x) = \frac{d}{dx}F(x)$.
+The definite integral is written:
+\begin{equation}
+\int_a^b f(x) dx= F(b) - F(a) \nonumber.
+\end{equation}
+
+Finally, we may see integration by parts in derivations. Recall that
+\begin{equation}
+\int u\,dv = uv + \int v\,du \nonumber.
+\end{equation}
+
+\section{Differentiation, too} 
+We will review plenty of differentiation when we discuss partial derivatives. We also write equations in a differential form and will solve simple differential equations. Be sure to review and practice the product rule and chain rule. A common mistake is to accidentally treat variables as constants and {\it vice versa}.
+\begin{center}
+\renewcommand{\arraystretch}{1.5}
+\setlength{\tabcolsep}{1em}
+\begin{tabular}{c}\\
+$\frac{d}{dx}(cu) = c\frac{du}{dx}$, where $c$ is a constant\\
+\end{tabular}
+\end{center}
+Contrast that to the case below when $u$ and $v$ are both functions of $x$. We must apply the \emph{product rule}:
+\begin{equation}
+\frac{d}{dx}(uv) = v\frac{du}{dx} + u\frac{dv}{dx} \nonumber
+\end{equation}
+
+Finally, review the other rules of differentiation, including the product rule and the chain rule,
+\begin{equation}
+\frac{d}{dx}(u + v) = \frac{du}{dx} + \frac{dv}{dx} \nonumber
+\end{equation}
+\begin{equation}
+\frac{d}{dx}\left ( \frac{u}{v} \right ) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \nonumber
+\end{equation}
+\begin{equation}
+\frac{d}{dx}u(v) = \frac{du}{dv}\frac{dv}{dx} \nonumber
+\end{equation}
+\begin{equation}
+\frac{1}{y}\frac{dy}{dx} = \frac{d \ln y}{dx} \nonumber
+\end{equation}
+\begin{equation}
+d\left ( \frac{1}{x} \right ) = - \frac{1}{x^2} dx \nonumber
+\end{equation}
+
+
+
+\section{Logarithmic functions} 
+Here, $\ln x$ is the natural logarithm ($\log_e x$). Most of the time when we see and use the function $\log$, we are referring to the logarithm with base 10, or $\log_{10}$. 
+\begin{center}
+\renewcommand{\arraystretch}{1.5}
+\setlength{\tabcolsep}{1em}
+\begin{tabular}{c}\\
+$\ln A + \ln B =$  $\ln (AB)$ \\
+$\ln A - \ln B =$  $\ln \frac{A}{B}$ \\
+$\ln x^a = a\ln x$\\
+%\end{tabular}
+%\begin{tabular}{c}\\
+$\ln e^x = x\ln e = x$ \\
+$\ln 1 = \ln e^0 = 0\times \ln e = 0$\\
+$\ln e = 1$\\
+\end{tabular}
+\end{center}
+
+\section{Exponential functions} 
+An exponential function is Euler's number $e$ raised to the $x$ power, $f(x) = e^x$. We've noted the relationship between the natural logarithm and $e$, which actually serves as a definition of $\ln x$: $\ln x = y \textrm{\ if and only if\ } e^y = x$. 
+\begin{center}
+\renewcommand{\arraystretch}{1.5}
+\setlength{\tabcolsep}{1em}
+\begin{tabular}{c}\\
+$e^0 = 1$\\
+$e^ae^b = e^{a+b}$\\
+$\frac{d}{dx}e^x = e^x$\\
+\end{tabular}
+\end{center}
+
+\section{Taylor series expansion} 
+The Taylor series is often used to linearize a function. Given the function $f(x)$, we can write
+\begin{equation}
+f(x) \approx f(x_0) + (x-x_0)f'(x_0) + ... \nonumber
+\end{equation}
+
+%\usepackage[style=unsrt, citestyle=unsrt]{biblatex}
+%\bibliography{references}
+
+\end{document}