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\hspace{-0.025\textwidth}\parbox[b]{1.05\textwidth}{\centering \bfseries
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{\Large \uppercase{CHEG 231 Chemical Engineering Thermodynamics}}\\
{\large {Department of Chemical and Biomolecular Engineering}}\\
{\large {University of Delaware}}
}

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{\uppercase{\Large Math reminders}\par}

%{\itshape Eric\ M.\ Furst\par}
{\itshape \today \par}

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%{\large Nome: \hrulefill\hspace{1em}{\small RA}: \rule{25mm}{0.4pt}}

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%\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.
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%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. 
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%\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 $\Gamma(x+1) = x!$}} \\
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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}.
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$\frac{d}{dx}(cu) = c\frac{du}{dx}$, where $c$ is a constant\\
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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 $\log_{10}$. 
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$\ln A + \ln B =$  $\ln (AB)$ \\
$\ln A - \ln B =$  $\ln \frac{A}{B}$ \\
$\ln x^a = a\ln x$\\
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$\ln e^x = x\ln e = x$ \\
$\ln 1 = \ln e^0 = 0\times \ln e = 0$\\
$\ln e = 1$\\
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\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 $\ln x$: $\ln x = y \textrm{\ if and only if\ } e^y = x$. 
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$e^0 = 1$\\
$e^ae^b = e^{a+b}$\\
$\frac{d}{dx}e^x = e^x$\\
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\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}

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