\begin{center}
\textbf{\Large Differentiation}
\end{center}
\definition{For a given function $f(x)$ the \textbf{derivative function} is given by
\[f'(x)=\mylim{h}{0}\frac{f(x+h)-f(x)}{h}~.\]
}
\begin{center}
\rule{.8\textwidth}{.05em}\\[1em]
\textbf{Table of Derivatives}
\end{center}
\begin{enumerate}
\item $\displaystyle\frac{d}{dx}(c)=0$
\item $\displaystyle\frac{d}{dx}(x^n)=nx^{n-1}$\hspace{8em} (\textbf{Power Rule})
\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 $\displaystyle \frac{d}{dx}\left(\csc x\right)=-\csc x\cot x$
\item $\displaystyle \frac{d}{dx}\left(\sec x\right)=\sec x\tan x$
\item $\displaystyle \frac{d}{dx}\left(\cot x\right)=-\csc^2 x$
\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{General Power Rule})
\item $\displaystyle\frac{d}{dx}f(g(x))=f'\left(g(x)\right)\cdot g'(x)$\hspace{3em} (\textbf{Chain Rule})
\end{enumerate}
\begin{center}
\rule{.8\textwidth}{.05em}
\end{center}
Here $c$ and $n$ are constants, $f$ and $g$ are functions, and primes ($f'$, $g'$) denote differentiation.

\newpage
\begin{center}
\textbf{\Large Integration}
\end{center}
\textbf{The Fundamental Theorem of Calculus (Antiderivative Form):}

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ (so $F'=f$) then
\[\int_a^bf(x)\, dx=F(b)-F(a)~.\]

\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}
\item $\displaystyle\int x^n\,dx=\frac{1}{n+1}x^{n+1}+C\ \ \ \ (n\neq-1)$
\item $\displaystyle\int \cos x\,dx=\sin x+C$
\item $\displaystyle\int \sin x\,dx=-\cos x+C$
\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 $\displaystyle\int \csc x\cot x\,dx=-\csc x+C$
\item $\displaystyle\int c f(x)\,dx=c\int f(x)\,dx$
\item $\displaystyle\int \left[f(x)\pm g(x)\right]\,dx=\int f(x)\,dx\pm \int g(x)\,dx$
\end{enumerate}
\begin{center}
\rule{.8\textwidth}{.05em}
\end{center}

\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.

\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
\[\int_a^b f(g(x))g'(x)\,dx=\int_{g(a)}^{g(b)} f(u)\,du\ .\]
