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% Document Info
\author{kristina\_hayes}
\pdfinfo{
  /Title (calculus-midterm-2.pdf)
  /Creator (Cheatography)
  /Author (kristina\_hayes)
  /Subject (Calculus Midterm 2 Cheat Sheet)
}

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\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
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    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Calculus Midterm 2 Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{kristina\_hayes} via \textcolor{DarkBackground}{\uline{cheatography.com/182718/cs/38026/}}}
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\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}kristina\_hayes \\
  \uline{cheatography.com/kristina-hayes} \\
  \end{tabulary}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 13th April, 2023.\\
   Page {\thepage} of \pageref{LastPage}.
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  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
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\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{x{3.13551 cm} x{1.84149 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Trigonometric Identities}}  \tn
% Row 0
\SetRowColor{LightBackground}
sin\textasciicircum{}2+cos\textasciicircum{}2=1 & sec(x) = 1/cos(x) \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
cot(x) = 1/tan(x)  OR  cos(x)/sin(x) & tan(x) = sin(x)/cos(x) \tn 
% Row Count 4 (+ 2)
% Row 2
\SetRowColor{LightBackground}
csc(x) = 1/sin(x) & sec\textasciicircum{}2 = tan\textasciicircum{}2+1 \tn 
% Row Count 6 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Graphing Steps}}  \tn
% Row 0
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\mymulticolumn{1}{x{5.377cm}}{1. Domain} \tn 
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% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{2. Intercepts} \tn 
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% Row 2
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\mymulticolumn{1}{x{5.377cm}}{3. Asymptotes} \tn 
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% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{4. Intervals of Increase and Decrease} \tn 
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% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{5. Local Minimums and Maximums} \tn 
% Row Count 5 (+ 1)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{6. Concavity and Inflection Points} \tn 
% Row Count 6 (+ 1)
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\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Graphing Tips}}  \tn
% Row 0
\SetRowColor{LightBackground}
VA: lim (x-\textgreater{}+\_infinity) f(x) =\_+infinity (left and right) & HA: lim (x-\textgreater{}+\_infinity) f(x) = c at y=c \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
VA: Find by setting the denominator = 0 and solving for x & HA: y=0 if n\textless{}d, ax/bx if n=d, none if n\textgreater{}d \tn 
% Row Count 6 (+ 3)
% Row 2
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First Derivative: Intervals of increase or decrease + min/max & Second Derivative: Concavity + Inflection Points \tn 
% Row Count 10 (+ 4)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
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\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.83689 cm} x{2.14011 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Derivative Rules}}  \tn
% Row 0
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Product: f'(x)g(x) + g'(x)f(x) & Chain: f'(g(x)) * g'(x) \tn 
% Row Count 2 (+ 2)
% Row 1
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\mymulticolumn{2}{x{5.377cm}}{Quotient: f'(x)g(x) - g'(x)f(x)/g(x)\textasciicircum{}2} \tn 
% Row Count 3 (+ 1)
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\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Acceleration and Velocity}}  \tn
% Row 0
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Acceleration is the antiderivative of velocity & (come to stop) 1. Find antiderivative of function \tn 
% Row Count 3 (+ 3)
% Row 1
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2. Find v(0) or C and set = 0 & (for distance) 1. Derivative, solve for t, derivative, plug in \tn 
% Row Count 7 (+ 4)
% Row 2
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\mymulticolumn{2}{x{5.377cm}}{To find t take derivative, to find distance take integral} \tn 
% Row Count 9 (+ 2)
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\begin{tabularx}{5.377cm}{x{2.28942 cm} x{2.68758 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Evaluating Integrals}}  \tn
% Row 0
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a+b/c = a/c + b/c & Indefinite: F(x) + C \tn 
% Row Count 1 (+ 1)
% Row 1
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\mymulticolumn{2}{x{5.377cm}}{F(b)-F(a) (find antiderivative and plug in)} \tn 
% Row Count 2 (+ 1)
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Unit Circle}}  \tn
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\mymulticolumn{1}{p{5.377cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/kristina-hayes_1680245576_UnitCircleAngles.jpg}}} \tn 
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\begin{tabularx}{5.377cm}{x{2.33919 cm} x{2.63781 cm} }
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\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Derivative Tests}}  \tn
% Row 0
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1st: Positive to Negative: local max & 2nd: f'(c) = 0 \& f''(c)\textgreater{}0: local min \& concave up \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
1st: Negative to Positive: local min & 2nd: f'(c) = 0 \& f''(c)\textless{}0: local max \& concave down \tn 
% Row Count 6 (+ 3)
% Row 2
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Critical points when f'(x)=0 & Inflection points when f''(x)=0 \tn 
% Row Count 8 (+ 2)
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\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.68758 cm} x{2.28942 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Intermediate Value Theorem}}  \tn
% Row 0
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a\textless{}c\textless{}b & Used to find when f(x) has roots \tn 
% Row Count 2 (+ 2)
% Row 1
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When proving roots, show that one part is positive and the other is negative & To find c, set y=0 and solve for x \tn 
% Row Count 6 (+ 4)
% Row 2
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\mymulticolumn{2}{x{5.377cm}}{To show at most, show that there is 1 critical value and f(x) can only cross x amount of times} \tn 
% Row Count 8 (+ 2)
% Row 3
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\mymulticolumn{2}{x{5.377cm}}{Explain that you are using IVT} \tn 
% Row Count 9 (+ 1)
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\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.58804 cm} x{2.38896 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Areas \& Distances}}  \tn
% Row 0
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Derivative: rate of change & Antiderivative: total change \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
n or change t = b-a/n & RHS: E (n i=1) f(ti) change t \tn 
% Row Count 4 (+ 2)
% Row 2
\SetRowColor{LightBackground}
LHS: E (n-1 i=0) f(ti) change t & ti = a +i change t \tn 
% Row Count 6 (+ 2)
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\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{U Subsitution}}  \tn
% Row 0
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\mymulticolumn{1}{x{5.377cm}}{Step 1: Make a "u-subsititution" (let u=)} \tn 
% Row Count 1 (+ 1)
% Row 1
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\mymulticolumn{1}{x{5.377cm}}{Step 2: Find du/dx} \tn 
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% Row 2
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\mymulticolumn{1}{x{5.377cm}}{Step 3: Solve for dx} \tn 
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% Row 3
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\mymulticolumn{1}{x{5.377cm}}{Step 4: Substitute dx and cancel out terms} \tn 
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% Row 4
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\mymulticolumn{1}{x{5.377cm}}{Step 5: Integrate with respect to u} \tn 
% Row Count 5 (+ 1)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{*If a definite integral, change the bounds from x bounds to u bounds} \tn 
% Row Count 7 (+ 2)
% Row 6
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\mymulticolumn{1}{x{5.377cm}}{*Add C if a indefinite integral} \tn 
% Row Count 8 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{3.88206 cm} x{1.09494 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Mean Value Theorem}}  \tn
% Row 0
\SetRowColor{LightBackground}
Is continuous and differentiable & \seqsplit{f(a)=f(b)} \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
f'(c)=f(b)-f(a)/b-a & f'(c)=0 \tn 
% Row Count 3 (+ 1)
% Row 2
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\mymulticolumn{2}{x{5.377cm}}{How large can this be?} \tn 
% Row Count 4 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{By MVT f'(c) =... for some c in {[}0,x{]}. Then do the math. Hence for every x in interval f(x) is whatever the math proves.} \tn 
% Row Count 7 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.9908 cm} x{2.9862 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Antiderivatives}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{Function}} & {\bf{Antiderivative}} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
x\textasciicircum{}n & x\textasciicircum{}n+1/n+1 \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
cos(x) & sin(x) \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
sin(x) & -cos(x) \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
sec\textasciicircum{}2(x) & tan(x) \tn 
% Row Count 5 (+ 1)
% Row 5
\SetRowColor{white}
sec(x)tan(x) & sec(x) \tn 
% Row Count 6 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.28942 cm} x{2.68758 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Derivatives}}  \tn
% Row 0
\SetRowColor{LightBackground}
{\bf{Fucntion}} & {\bf{Derivative}} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
sin(x) & cos(x) \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
cos(x) & -sin(x) \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
tan(x) & sec\textasciicircum{}2(x) \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
csc(x) & -csc(x) \tn 
% Row Count 5 (+ 1)
% Row 5
\SetRowColor{white}
sec(x) & sec(x)tan(x) \tn 
% Row Count 6 (+ 1)
% Row 6
\SetRowColor{LightBackground}
cot(x) & -csc\textasciicircum{}2(x) \tn 
% Row Count 7 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Optimization Problems}}  \tn
% Row 0
\SetRowColor{LightBackground}
Usually using two different formulas (like volume and perimeter) & If maximizing volume, solve for one variable and plug that it \tn 
% Row Count 4 (+ 4)
% Row 1
\SetRowColor{white}
Next, solve for derivative and set = 0 & After solving for that variable, plug into original (volume) equation \tn 
% Row Count 8 (+ 4)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{For distance: √(x-a)\textasciicircum{}2 + (y-b)\textasciicircum{}2 \& solve for critical point} \tn 
% Row Count 10 (+ 2)
% Row 3
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{May need to prove that something is a global min/max} \tn 
% Row Count 12 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{p{0.4977 cm} p{0.4977 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Properties of the Definite Integral}}  \tn
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\mymulticolumn{2}{x{5.377cm}}{Constant:} \tn 
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% Row 1
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\mymulticolumn{2}{x{5.377cm}}{Addition:} \tn 
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% Row 2
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\mymulticolumn{2}{x{5.377cm}}{Pulling a Constant:} \tn 
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% Row 3
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{Subtraction} \tn 
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% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{Splitting} \tn 
% Row Count 5 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}