\documentclass[10pt,a4paper]{article}

% Packages
\usepackage{fancyhdr}           % For header and footer
\usepackage{multicol}           % Allows multicols in tables
\usepackage{tabularx}           % Intelligent column widths
\usepackage{tabulary}           % Used in header and footer
\usepackage{hhline}             % Border under tables
\usepackage{graphicx}           % For images
\usepackage{xcolor}             % For hex colours
%\usepackage[utf8x]{inputenc}    % For unicode character support
\usepackage[T1]{fontenc}        % Without this we get weird character replacements
\usepackage{colortbl}           % For coloured tables
\usepackage{setspace}           % For line height
\usepackage{lastpage}           % Needed for total page number
\usepackage{seqsplit}           % Splits long words.
%\usepackage{opensans}          % Can't make this work so far. Shame. Would be lovely.
\usepackage[normalem]{ulem}     % For underlining links
% Most of the following are not required for the majority
% of cheat sheets but are needed for some symbol support.
\usepackage{amsmath}            % Symbols
\usepackage{MnSymbol}           % Symbols
\usepackage{wasysym}            % Symbols
%\usepackage[english,german,french,spanish,italian]{babel}              % Languages

% Document Info
\author{Mac1223 (Annie1223)}
\pdfinfo{
  /Title (rules-of-differentiation-and-comp-statics-use.pdf)
  /Creator (Cheatography)
  /Author (Mac1223 (Annie1223))
  /Subject (Rules of Differentiation \& Comp. Statics Use Cheat Sheet)
}

% Lengths and widths
\addtolength{\textwidth}{6cm}
\addtolength{\textheight}{-1cm}
\addtolength{\hoffset}{-3cm}
\addtolength{\voffset}{-2cm}
\setlength{\tabcolsep}{0.2cm} % Space between columns
\setlength{\headsep}{-12pt} % Reduce space between header and content
\setlength{\headheight}{85pt} % If less, LaTeX automatically increases it
\renewcommand{\footrulewidth}{0pt} % Remove footer line
\renewcommand{\headrulewidth}{0pt} % Remove header line
\renewcommand{\seqinsert}{\ifmmode\allowbreak\else\-\fi} % Hyphens in seqsplit
% This two commands together give roughly
% the right line height in the tables
\renewcommand{\arraystretch}{1.3}
\onehalfspacing

% Commands
\newcommand{\SetRowColor}[1]{\noalign{\gdef\RowColorName{#1}}\rowcolor{\RowColorName}} % Shortcut for row colour
\newcommand{\mymulticolumn}[3]{\multicolumn{#1}{>{\columncolor{\RowColorName}}#2}{#3}} % For coloured multi-cols
\newcolumntype{x}[1]{>{\raggedright}p{#1}} % New column types for ragged-right paragraph columns
\newcommand{\tn}{\tabularnewline} % Required as custom column type in use

% Font and Colours
\definecolor{HeadBackground}{HTML}{333333}
\definecolor{FootBackground}{HTML}{666666}
\definecolor{TextColor}{HTML}{333333}
\definecolor{DarkBackground}{HTML}{40A37F}
\definecolor{LightBackground}{HTML}{F3F9F7}
\renewcommand{\familydefault}{\sfdefault}
\color{TextColor}

% Header and Footer
\pagestyle{fancy}
\fancyhead{} % Set header to blank
\fancyfoot{} % Set footer to blank
\fancyhead[L]{
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
    \SetRowColor{DarkBackground}
    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
    }
\end{tabulary}
\columnbreak
\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Rules of Differentiation \& Comp. Statics Use Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{Mac1223 (Annie1223)} via \textcolor{DarkBackground}{\uline{cheatography.com/135483/cs/28136/}}}
\end{tabulary}
\end{multicols}}

\fancyfoot[L]{ \footnotesize
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
  \SetRowColor{FootBackground}
  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}Mac1223 (Annie1223) \\
  \uline{cheatography.com/annie1223} \\
  \end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 27th May, 2021.\\
   Updated 27th May, 2021.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
  \SetRowColor{white}
  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




\begin{document}
\raggedright
\raggedcolumns

% Set font size to small. Switch to any value
% from this page to resize cheat sheet text:
% www.emerson.emory.edu/services/latex/latex_169.html
\footnotesize % Small font.

\begin{multicols*}{2}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Rules of Differentiation for a fn. of 1 Variable}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{Constant Function Rule}} \newline % Row Count 1 (+ 1)
y = f({\emph{x}}) = k \newline % Row Count 2 (+ 1)
d/dy(f({\emph{x}})) = d/dy(k) = 0 \newline % Row Count 3 (+ 1)
{\bf{Power Rule}} \newline % Row Count 4 (+ 1)
y = f({\emph{x}}) = x\textasciicircum{}n\textasciicircum{} \newline % Row Count 5 (+ 1)
d/dy(f({\emph{x}})) = d/dy(x\textasciicircum{}n\textasciicircum{}) = n.x\textasciicircum{}n-1\textasciicircum{} \newline % Row Count 6 (+ 1)
{\bf{Power Function Rule Generalized}} \newline % Row Count 7 (+ 1)
y = f({\emph{x}}) = cx\textasciicircum{}n\textasciicircum{} (where c is the multiplicative constant \newline % Row Count 9 (+ 2)
d/dy(f({\emph{x}})) = d/dy(cx\textasciicircum{}n\textasciicircum{}) = cnx\textasciicircum{}n-1\textasciicircum{}% Row Count 10 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Rules of Diff. for 2 or \textgreater{} fn. of the same variable}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{Sum \& Diff Rule}} \newline % Row Count 1 (+ 1)
 d/dy {[}f({\emph{x}}) ± g({\emph{x}}){]} = d/dy {[}f({\emph{x}}){]} ± d/dy {[}g({\emph{x}}){]} = f '(x) ± g'(x) \newline % Row Count 3 (+ 2)
{\bf{Product Rule}} \newline % Row Count 4 (+ 1)
 d/dy {[}f({\emph{x}}).g({\emph{x}}){]} = f({\emph{x}}).d/dy {[}g({\emph{x}}){]} + g({\emph{x}}).d/dy {[}f({\emph{x}}){]} = f({\emph{x}}).g'(x) + g({\emph{x}}).f '(x) \newline % Row Count 6 (+ 2)
{\bf{Finding Marginal Revenue Function from Average Revenue Function}} \newline % Row Count 8 (+ 2)
AR = 15 - Q, TR = AR.Q = 15Q-Q\textasciicircum{}2\textasciicircum{}, MR = 15 - 2Q \newline % Row Count 9 (+ 1)
AR = TR/Q = P.Q/Q = P , therefore {[}Average Revenue is the inverse of the demand as AR is the curve relating Price to output: P = f(Q){]} \newline % Row Count 12 (+ 3)
Under Perfect Competition, the AR curve is horizontal straight line ( MR - AR = ), therefore they must coincide). Under imperfect competition, AR is a downward sloping curve (where MR - AR \textless{} 0 for all positive levels of output, MR is above AR) \newline % Row Count 17 (+ 5)
{\bf{Quotient Rule}} \newline % Row Count 18 (+ 1)
d/dy {[}f({\emph{x}})/g({\emph{x}}){]} =  f '(x).g({\emph{x}}) + g'(x).f({\emph{x}})/{[}g({\emph{x}}){]}\textasciicircum{}2\textasciicircum{} \newline % Row Count 20 (+ 2)
{\bf{Relationship Between Marginal Cost \& Average Cost Functions}} \newline % Row Count 22 (+ 2)
AC = C(Q)/Q as long as Q\textgreater{}0, The rate of change of AC WRT Q is done by differentiating AC. \newline % Row Count 24 (+ 2)
AC' = 1/Q {[} C'(Q) - C(Q)/Q{]}, MC = C'(Q) \newline % Row Count 25 (+ 1)
{\bf{Exercise 1}}: \newline % Row Count 26 (+ 1)
A firm produces 'q' units of water. Its total revenue and total cost functions are given as: \newline % Row Count 28 (+ 2)
TR=-0.5q\textasciicircum{}2\textasciicircum{}+100q and TC=5q\textasciicircum{}2\textasciicircum{}+12q+200 \newline % Row Count 29 (+ 1)
a) Calculate MR and interpret your solution. \newline % Row Count 30 (+ 1)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Rules of Diff. for 2 or \textgreater{} fn. of the same variable (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{MR = dTR/dQ = -q + 100 {[}It is a decreasing function, therefore a downward sloping curve, MR is the extra revenue that an additional unit of q will bring to the firm, when q = 0, MR is 100 implying that an additional unit of q will decrease the extra revenue{]} \newline % Row Count 6 (+ 6)
b) Calculate MC and interpret your solution. \newline % Row Count 7 (+ 1)
MC = dTC/dQ = 10q + 12 {[}It is an increasing function, therefore a upward sloping curve, MC is the extra cost that an additional unit of q will bring to the firm, when q = 0, MC = 12 implying that an additional unit of q will increase the extra cost{]} \newline % Row Count 12 (+ 5)
c) How many units of water should the firm produce to maximize total profit? \newline % Row Count 14 (+ 2)
{\emph{Profit Max = MR = MC}} \newline % Row Count 15 (+ 1)
-q+100=10q+12 \newline % Row Count 16 (+ 1)
11q=88 \newline % Row Count 17 (+ 1)
q=8% Row Count 18 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Partial Differentiation}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{Partial Derivatives}} \newline % Row Count 1 (+ 1)
{\emph{f}} 1 = ∂y/∂{\emph{x}}1, this is the partial derivative WRT {\emph{x}}1, all other independent variables in function are held constant \newline % Row Count 4 (+ 3)
{\bf{Techniques of Partial Differentiation}} \newline % Row Count 5 (+ 1)
1.{\emph{Simple Rule}} \newline % Row Count 6 (+ 1)
 y = f(X1,X2) = 3X\textasciicircum{}2\textasciicircum{}1 + X1X2 + 4X\textasciicircum{}2\textasciicircum{}2 \newline % Row Count 7 (+ 1)
∂y/∂{\emph{x}}1 = 6X1+X2; ∂y/∂{\emph{x}}2 = X1+8X2 \newline % Row Count 8 (+ 1)
2. {\emph{Product Rule}} \newline % Row Count 9 (+ 1)
 y = f(u,v) = (u+4)(3u+2v) \newline % Row Count 10 (+ 1)
fu = (u+4)(3) + (3u+2v)(1) = 6u + 2v + 12 \newline % Row Count 11 (+ 1)
fv = (u+4)(2) + (3u+2v)(0) = 2u+8 \newline % Row Count 12 (+ 1)
3.{\emph{Quotient Rule}} \newline % Row Count 13 (+ 1)
y = f(u,v) = (3u-2v)/(u\textasciicircum{}2\textasciicircum{} + 3v) \newline % Row Count 14 (+ 1)
fu = {[}(3)(u\textasciicircum{}2\textasciicircum{} + 3v) - (2u)(3u-2v){]}/{[}(u\textasciicircum{}2\textasciicircum{} + 3v){]}\textasciicircum{}2\textasciicircum{} {\emph{simplify it}} \newline % Row Count 16 (+ 2)
fv = {[}(-2)(u\textasciicircum{}2\textasciicircum{} + 3v) - (3)(3u-2v){]}/{[}(u\textasciicircum{}2\textasciicircum{} + 3v){]}\textasciicircum{}2\textasciicircum{}  {\emph{simplify it}} \newline % Row Count 18 (+ 2)
A partial derivative is a measure of the instantaneous rates of change of some variable. \newline % Row Count 20 (+ 2)
MPPk = Marginal physical Product of Capital, which is the partial derivative Qk relates to the changes of output WRT infinitesimal changes in capital while labour input is held constant. Similar the partial Derivative of Ql is the mathematical representation of the MPPL function. \newline % Row Count 26 (+ 6)
{\bf{Exercise 4}}:  \newline % Row Count 27 (+ 1)
Given the following utility function:  f(x,y,z)=x\textasciicircum{}2\textasciicircum{} y\textasciicircum{}3\textasciicircum{} z    0\textless{}x,y,z\textless{}1 \newline % Row Count 29 (+ 2)
a) Calculate marginal utility with respect to y and interpret. \newline % Row Count 31 (+ 2)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Partial Differentiation (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{∂U/∂y = 3x\textasciicircum{}2\textasciicircum{}y\textasciicircum{}2\textasciicircum{}z {[}MU is + \& \textgreater{} 0 for all values of x,y,z, function is strictly monotonic{]}% Row Count 2 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Rules of Diff. with fns. of Different Variables}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{Chain Rule}} \newline % Row Count 1 (+ 1)
z = f(y), y = g(x); dz/dx = (dz/dy).(dy/dx) = f '(y). g'(x) \newline % Row Count 3 (+ 2)
{\emph{function can extend to 3 variables}}:  \newline % Row Count 4 (+ 1)
z = f(y), y = g(x), x = h(w) \newline % Row Count 5 (+ 1)
dz/dw = (dz/dy).(dy/dx).(dx/dw) = f '(y). g'(x). h '(w) \newline % Row Count 7 (+ 2)
Given that TR = f(Q), where Q is a function of labour input, Q = g(L), find dTR/dL, by chain rule, we get dTR/dL = (dTR/dQ).(dQ/dL) = f '(Q).g'(L), {[}dTR/dQ is the MR function and dQ/dL is the \seqsplit{marginal-physical-product} of Labour (MPPL) function. dTR/dL has the connotation of the Marginal revenue product of labour (MRPL) \newline % Row Count 14 (+ 7)
{\bf{Exercise 2}}:  \newline % Row Count 15 (+ 1)
Given TR= PQ and Q=A.K\textasciicircum{}0.5\textasciicircum{}L\textasciicircum{}0.3\textasciicircum{}H\textasciicircum{}0.2\textasciicircum{} \newline % Row Count 16 (+ 1)
a) Calculate ∂TR/∂K. Interpret your solution. \newline % Row Count 17 (+ 1)
∂TR/∂K = (dTR/dQ).(dQ/dK) = P.{[}A.0.5.K\textasciicircum{}-0.5\textasciicircum{}.L\textasciicircum{}0.3\textasciicircum{}H\textasciicircum{}0.2\textasciicircum{}{]}  \newline % Row Count 19 (+ 2)
= 0.5.P.A.L\textasciicircum{}0.3\textasciicircum{}H\textasciicircum{}0.2\textasciicircum{}/K\textasciicircum{}0.5\textasciicircum{}, this is the Marginal Revenue Product of Capital, how does total revenue change when an extra unit of output is produced when one unit of capital is added, dQ/dK is the marginal product of capital (MPK). \newline % Row Count 24 (+ 5)
b) Calculate ∂Q/∂H. Interpret your solution. \newline % Row Count 25 (+ 1)
∂Q/∂H = 0.2 A.K\textasciicircum{}0.5\textasciicircum{}L\textasciicircum{}0.3\textasciicircum{}/H\textasciicircum{}0.8\textasciicircum{}, this is the marginal product of Human Capital, it is the extra amount of output that is produced when one unit of Human capital is added, holding all the other inputs constant. \newline % Row Count 30 (+ 5)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Rules of Diff. with fns. of Different Variables (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{Inverse Function Rule}} \newline % Row Count 1 (+ 1)
If a function y=f(x) represents a one-to-one mapping (for every x value, it has a unique y value, vice versa) then f has an inverse {\emph{f}} \textasciicircum{}-1\textasciicircum{}. \newline % Row Count 4 (+ 3)
A function is strictly increasing if: \newline % Row Count 5 (+ 1)
X1\textgreater{}X2 -{}-\textgreater{} f(X1) \textgreater{} f(X2) for all values of  x, y  \newline % Row Count 6 (+ 1)
A function is strictly decreasing if: \newline % Row Count 7 (+ 1)
X1\textgreater{}X2 -{}-\textgreater{} f(X1) \textless{} f(X2) for all values of x, y \newline % Row Count 8 (+ 1)
If a function is either strictly increasing or strictly decreasing it is strictly monotonic. \newline % Row Count 10 (+ 2)
If a function is monotonic then the function has an inverse. \newline % Row Count 12 (+ 2)
Thus for example, firms demand curve Q = f(P) that has a negative slope throughout is decreasing, as such it has an inverse function, P = f \textasciicircum{}-1\textasciicircum{} (Q) which gives the Av. revenue of the firm P = AR. \newline % Row Count 16 (+ 4)
For inverse functions, the rule of differentiation: dx/dy = 1/{[}dy/dx{]} \newline % Row Count 18 (+ 2)
{\emph{check for strictly monotonic: f'(x) \textgreater{} 0 or f'(x) \textless{}0 for all x.}} If function is a U-shaped curve, may have to check which part of the curve is increasing and which part is decreasing for the x values. \newline % Row Count 23 (+ 5)
{\bf{Exercise 3}}: \newline % Row Count 24 (+ 1)
Using the TR and TC functions from Exercise 1: \newline % Row Count 25 (+ 1)
a)	Determine whether the total cost function is strictly increasing? \newline % Row Count 27 (+ 2)
TC=5q\textasciicircum{}2\textasciicircum{}+12q+200, if q\textgreater{}0 then TC will increase \newline % Row Count 28 (+ 1)
MC = 10q+12, it will be positive for all values of q if q\textgreater{}0 \newline % Row Count 30 (+ 2)
} \tn 
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Rules of Diff. with fns. of Different Variables (cont)}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{b)	For what values of q will the total revenue function be strictly increasing? \newline % Row Count 2 (+ 2)
TR=-0.5q\textasciicircum{}2\textasciicircum{}+100q \newline % Row Count 3 (+ 1)
MR=-q+100 (when q = 0 , MR = 100 which is the turning point for TR) therfore TR will be increasing for q values that are \textless{} 100% Row Count 6 (+ 3)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Application to Comparative Static Analysis}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{Market Model}} \newline % Row Count 1 (+ 1)
Look at page 170 - 172 in textbook \newline % Row Count 2 (+ 1)
{\bf{National - Income Model}} \newline % Row Count 3 (+ 1)
Look at page 172 - 173 in textbook, it goes over the income equilibrium and the partial derivative to get the government expenditure multiplier, non-income tax multiplier and how an increase in the income tax rate will lower equilibrium income.% Row Count 8 (+ 5)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Note on Jacobian Determinants}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{8.4cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/annie1223_1622140094_7.6.PNG}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{The two equations are linearly dependent if ǀJǀ = 0}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


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

\end{document}