\begin{center}
%\rule{.8\textwidth}{.05em}\\[1em]
\textbf{Table of Derivatives}
\end{center}
\begin{enumerate}
  \begin{multicols}{2}
\item $\displaystyle\frac{d}{dx}(c)=0$
\item $\displaystyle\frac{d}{dx}(x^n)=nx^{n-1}$%\hspace{8em} (\textbf{Power Rule})
\end{multicols}
\rule{.8\textwidth}{.05em}
\begin{multicols}{2}
\item $\displaystyle \frac{d}{dx}\left(e^x\right)=e^x$
\item $\displaystyle \frac{d}{dx}\left(a^x\right)=a^x\ln a$
\item $\displaystyle \frac{d}{dx}\left(\ln x\right)=\frac{1}{x}$
\item $\displaystyle \frac{d}{dx}\left(\log_a x\right)=\frac{1}{x\ln a}$
\end{multicols}
\rule{.8\textwidth}{.05em}
\begin{multicols}{2}
\item $\displaystyle \frac{d}{dx}\left(\sin x\right)=\cos x$
\item $\displaystyle \frac{d}{dx}\left(\cos x\right)=-\sin x$
\item $\displaystyle \frac{d}{dx}\left(\tan x\right)=\sec^2 x$
\item $\displaystyle \frac{d}{dx}\left(\sec x\right)=\sec x\tan x$
\item $\displaystyle \frac{d}{dx}\left(\csc x\right)=-\csc x\cot x$
\item $\displaystyle \frac{d}{dx}\left(\cot x\right)=-\csc^2 x$
\end{multicols}
\rule{.8\textwidth}{.05em}
\begin{multicols}{2}
\item $\displaystyle\frac{d}{dx}\left(\sin^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}$
\item $\displaystyle\frac{d}{dx}\left(\cos^{-1}x\right)=-\frac{1}{\sqrt{1-x^2}}$
\item $\displaystyle\frac{d}{dx}\left(\tan^{-1}x\right)=\frac{1}{1+x^2}$
\item $\displaystyle\frac{d}{dx}\left(\sec^{-1}x\right)=\frac{1}{x\sqrt{x^2-1}}$
\item $\displaystyle\frac{d}{dx}\left(\csc^{-1}x\right)=-\frac{1}{x\sqrt{x^2-1}}$
\item $\displaystyle\frac{d}{dx}\left(\cot^{-1}x\right)=-\frac{1}{1+x^2}$
\end{multicols}
\rule{.8\textwidth}{.05em}
\item $\displaystyle \frac{d}{dx}\left(cf\right)=c\frac{df}{dx}$
\item $\displaystyle \frac{d}{dx}\left(f\pm g\right)=\frac{df}{dx}\pm\frac{dg}{dx}$
\item $\displaystyle \frac{d}{dx}\left(fg\right)=\frac{df}{dx}g+f\frac{dg}{dx}$\hspace{5.5em} (\textbf{Product Rule})
\item $\displaystyle \frac{d}{dx}\left(\frac{f}{g}\right)=\frac{\frac{df}{dx}g-f\frac{dg}{dx}}{g^2}$\hspace{5.4em} (\textbf{Quotient Rule})
\item $\displaystyle\frac{d}{dx}\left[f(x)\right]^n=n\left[f(x)\right]^{n-1}f'(x)$\hspace{2.2em} (\textbf{Generalized Power Rule})
\item $\displaystyle\frac{d}{dx}f(g(x))=f'\left(g(x)\right)\cdot g'(x)$\hspace{3em} (\textbf{Chain Rule})
  \\[1em]
\rule{.8\textwidth}{.05em}
\end{enumerate}

Here $c$, $n$, and $a>0$  are constants, $f$ and $g$ are functions of $x$, and primes ($f'$, $g'$) denote differentiation.
\newpage
%\begin{center}
%\textbf{Integration}
%\end{center}
%\theorem{The Fundamental Theorem of Calculus (Part II):
%\begin{description}
%\item{\textbf{Part I:}} Suppose $f$ is continuous on $[a,b]$, then the function $g$ defined by
%\[g(x)=\int_a^xf(t)\,dt\ \ \ ;\ \ \ a\le x\le b,\]
%is continuous on $[a,b]$ and differentiable on $(a,b)$ with
%\[g'(x)=f(x)\ ,\]
%or, equivalently,
%\[\frac{d}{dx}\int_a^xf(t)\,dt=f(x)\ .\]
%\item{\textbf{Part II:}}

% If $f$ is continuous on $[a,b]$ then\
%\[\int_a^bf(x)\, dx=F(b)-F(a)=\left. F(x)\right|_a^b\]
%for any $F$ an antiderivative of $f$, i.e. $F'(x)=f(x)$.
%%\end{description}
%}

%\definition{If $F(x)$ is an antiderivative of $f$, so $F'(x)=f(x)$, then the \textbf{indefinite integral} of $f(x)$ is
%\[\int f(x)\,dx=F(x)+C\]
%}

\begin{center}
%\rule{.8\textwidth}{.05em}\\[1em]
\textbf{Table of Indefinite Integrals}
\end{center}
\begin{enumerate}
\begin{multicols}{2}
\item $\displaystyle\int x^n\,dx=\frac{1}{n+1}x^{n+1}+C\ \ \ \ (n\neq-1)$
\item $\displaystyle\int\frac{1}{x}\,dx=\ln|x|+C$
\end{multicols}
\rule{.8\textwidth}{.05em}
\begin{multicols}{2}
\item $\displaystyle\int e^x\,dx=e^x+C$
%\item $\displaystyle\int e^{ax}\,dx=\frac{1}{a}e^{ax}+C$
\item $\displaystyle\int a^x\,dx=\frac{1}{\ln a}a^x+C$
\end{multicols}
\rule{.8\textwidth}{.05em}
\begin{multicols}{2}
\item $\displaystyle\int \sin x\,dx=-\cos x+C$
\item $\displaystyle\int \cos x\,dx=\sin x+C$
\item $\displaystyle\int \tan x\,dx=\ln|\sec x|+C$
\item $\displaystyle\int \sec x\,dx=\ln|\sec x+\tan x|+C$
\item $\displaystyle\int \csc x\,dx=\ln|\csc x-\cot x|+C$
\item $\displaystyle\int \cot x\,dx=\ln|\sin x|+C$
\end{multicols}
\rule{.8\textwidth}{.05em}
\begin{multicols}{2}
\item $\displaystyle\int \sec^2x\,dx=\tan x+C$
\item $\displaystyle\int \sec x\tan x\,dx=\sec x+C$
\item $\displaystyle\int \csc^2x \,dx=-\cot x+C$
\item $\displaystyle\int \csc x\cot x\,dx=-\csc x+C$
\end{multicols}
\rule{.8\textwidth}{.05em}
\begin{multicols}{2}
\item $\displaystyle\int\frac{1}{\sqrt{1-x^2}}\,dx=\sin^{-1}x+C$
\item $\displaystyle\int\frac{1}{1+x^2}\,dx=\tan^{-1}x+C$
\item $\displaystyle\int\frac{1}{x\sqrt{x^2-1}}\,dx=\sec^{-1}x+C$
\item $\displaystyle\int\frac{1}{\sqrt{a^2-x^2}}\,dx=\sin^{-1}\frac{x}{a}+C$%\ \ \ \ (a>0)$
\item $\displaystyle\int\frac{1}{a^2+x^2}\,dx=\frac{1}{a}\tan^{-1}\frac{x}{a}+C$
\item $\displaystyle\int\frac{1}{x\sqrt{x^2-a^2}}\,dx=\frac{1}{a}\sec^{-1}\frac{x}{a}+C$
\end{multicols}
\rule{.8\textwidth}{.05em}
%\begin{multicols}{2}
\item $\displaystyle\int c f(x)\,dx=c\int f(x)\,dx$
\item $\displaystyle\int \left[f(x)\pm g(x)\right]\,dx=\int f(x)\,dx\pm \int g(x)\,dx$
\item $\displaystyle \int f(g(x))g'(x)\,dx=\int f(u)\,du\ \text{where}\ u=g(x)$ \hspace{3em} (\textbf{Substitution Rule})
\item $\displaystyle \int u\,dv=uv-\int v\,du$ \hspace{13em} (\textbf{Integration by Parts})\\[1em]
\rule{.8\textwidth}{.05em}
\end{enumerate}


Here $c$, $n$, and $a>0$  are constants, $f$ and $g$ are functions of $x$, and primes ($g'$) denote differentiation.

%\theorem{\textbf{Substitution Rule (Indefinite Integrals):} Suppose $u=g(x)$ is a differentiable function whose range of values is an interval $I$ upon which a further function $f$ is continuous, then
%\[\int f(g(x))g'(x)\,dx=\int f(u)\,du\ ,\]
%where $du=g'(x)dx$ and the right hand integral is to be evaluated at $u=g(x)$ after integration.
%}
%
%Note that the $du$ appearing on the right side is the differential:
%\[du=g'(x)dx\]
%
%\theorem{\textbf{Substitution Rule (Definite Integrals):} Suppose
%  $u=g(x)$ is a differentiable function whose derivative $g'$ is
%  continuous on $[a,b]$ and a further function $f$ is continuous on
%  the range of $u=g(x)$ (evaluated on $[a,b]$), then
%\[\int_a^b f(g(x))g'(x)\,dx=\int_{g(a)}^{g(b)} f(u)\,du\ .\]
%}
