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\author{Zniv}
\pdfinfo{
  /Title (basic-calculus-derivative-of-trig-funcs.pdf)
  /Creator (Cheatography)
  /Author (Zniv)
  /Subject (Basic Calculus Derivative of Trig Funcs Cheat Sheet)
}

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\noindent
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    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
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    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Basic Calculus Derivative of Trig Funcs Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{Zniv} via \textcolor{DarkBackground}{\uline{cheatography.com/134254/cs/43319/}}}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}Zniv \\
  \uline{cheatography.com/zniv} \\
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 6th May, 2024.\\
   Page {\thepage} of \pageref{LastPage}.
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% Row 0
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\mymulticolumn{1}{x{3.833cm}}{{\bf{Reciprocal Trigonometric Identities}}} \tn 
% Row Count 1 (+ 1)
% Row 1
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\mymulticolumn{1}{x{3.833cm}}{Sin θ = 1/Csc θ or Csc θ = 1/Sin θ} \tn 
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% Row 2
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\mymulticolumn{1}{x{3.833cm}}{Cos θ = 1/Sec θ or Sec θ = 1/Cos θ} \tn 
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\mymulticolumn{1}{x{3.833cm}}{Tan θ = 1/Cot θ or Cot θ = 1/Tan θ} \tn 
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% Row 1
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\mymulticolumn{1}{x{3.833cm}}{We will determine the derivative of sec x using the chain rule. We will use the following formulas and identities to calculate the derivative: \newline % Row Count 3 (+ 3)
sec x = 1/cos x \newline % Row Count 4 (+ 1)
tan x = sin x/cos x \newline % Row Count 5 (+ 1)
(cos x)' = -sin x \newline % Row Count 6 (+ 1)
(sec x)' = (1/cos x)' = (-1/cos2x).(cos x)' \newline % Row Count 7 (+ 1)
= (-1/cos2x).(-sin x) \newline % Row Count 8 (+ 1)
= sin x/cos2x \newline % Row Count 9 (+ 1)
= (sin x/cos x).(1/cos x) \newline % Row Count 10 (+ 1)
= tan x sec x \newline % Row Count 11 (+ 1)
Therefore, d(sec x)/dx = tan x sec x% Row Count 12 (+ 1)
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\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{We will determine the derivative of cot x using the quotient rule. We will use the following formulas and identities to calculate the derivative: \newline % Row Count 3 (+ 3)
(sin x)' = cos x \newline % Row Count 4 (+ 1)
(cos x)' = -sin x \newline % Row Count 5 (+ 1)
cot x = cos x/ sin x \newline % Row Count 6 (+ 1)
cos2x + sin2x = 1 \newline % Row Count 7 (+ 1)
cosec x = 1/sin x \newline % Row Count 8 (+ 1)
(cot x)' = (cos x/sin x)' \newline % Row Count 9 (+ 1)
= {[}(cos x)' sin x - (sin x)' cos x{]}/sin2x \newline % Row Count 10 (+ 1)
= {[}-sin x. sin x - cos x. cos x{]}/sin2x \newline % Row Count 11 (+ 1)
= (-sin2x - cos2x)/sin2x \newline % Row Count 12 (+ 1)
= -1/sin2x \newline % Row Count 13 (+ 1)
= -cosec2x \newline % Row Count 14 (+ 1)
Therefore, d(cot x)/dx = -cosec2x% Row Count 15 (+ 1)
} \tn 
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\begin{tabularx}{3.833cm}{X}
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\mymulticolumn{1}{x{3.833cm}}{\bf\textcolor{white}{Derivative of cosec x}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{3.833cm}}{We will determine the derivative of cosec x using the chain rule. We will use the following formulas and identities to calculate the derivative: \newline % Row Count 3 (+ 3)
cosec x = 1/sin x \newline % Row Count 4 (+ 1)
cot x = cos x/sin x \newline % Row Count 5 (+ 1)
(sin x)' = cos x \newline % Row Count 6 (+ 1)
(cosec x)' = (1/sin x)' = (-1/sin2x).(sin x)' \newline % Row Count 7 (+ 1)
= (-1/sin2x).(cos x) \newline % Row Count 8 (+ 1)
= -cos x/sin2x \newline % Row Count 9 (+ 1)
= -(cos x/sin x).(1/sin x) \newline % Row Count 10 (+ 1)
= -cot x cosec x \newline % Row Count 11 (+ 1)
Therefore, d(cosec x)/dx = -cot x cosec x% Row Count 12 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
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% That's all folks
\end{multicols*}

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