Generate Math Exams and Quizzes in LaTeX and pdf formats

Identifying \(\displaystyle f=8 x + 2\) and \(\displaystyle g=\sin{\left(x \right)} \cos{\left(x \right)}\) and using the product rule with \(\displaystyle f=8 x + 2 \implies f'=8\). This leaves g as \(\displaystyle g = \sin{\left(x \right)} \cos{\left(x \right)}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)}\) and \(\displaystyle g=\cos{\left(x \right)} \implies g'=- \sin{\left(x \right)}\). Popping up a level gives \(\displaystyle g'=(\cos{\left(x \right)})(\cos{\left(x \right)})+(\sin{\left(x \right)})(- \sin{\left(x \right)})\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(8 x + 2)(- \sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)})+(\sin{\left(x \right)} \cos{\left(x \right)})(8)=- \left(8 x + 2\right) \sin^{2}{\left(x \right)} + \left(8 x + 2\right) \cos^{2}{\left(x \right)} + 8 \sin{\left(x \right)} \cos{\left(x \right)}\)

Download \(\LaTeX\)

\begin{question}Find the derivative of $y = (8 x + 2)(\sin{\left(x \right)})(\cos{\left(x \right)})$.
    \soln{9cm}{Identifying $f=8 x + 2$ and $g=\sin{\left(x \right)} \cos{\left(x \right)}$ and using the product rule with $f=8 x + 2 \implies f'=8$. This leaves g as $g = \sin{\left(x \right)} \cos{\left(x \right)}$ which also requires the product rule. Pushing down in the new product rule $f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)}$ and $g=\cos{\left(x \right)} \implies g'=- \sin{\left(x \right)}$. Popping up a level gives $g'=(\cos{\left(x \right)})(\cos{\left(x \right)})+(\sin{\left(x \right)})(- \sin{\left(x \right)})$Popping up again (Back to the original problem) gives $f'=(8 x + 2)(- \sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)})+(\sin{\left(x \right)} \cos{\left(x \right)})(8)=- \left(8 x + 2\right) \sin^{2}{\left(x \right)} + \left(8 x + 2\right) \cos^{2}{\left(x \right)} + 8 \sin{\left(x \right)} \cos{\left(x \right)}$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)

\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}

\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}

\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}

\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue

\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

\end{question}\end{document}

HTML for Canvas

<p> <p>Find the derivative of  <img class="equation_image" title=" \displaystyle y = (8 x + 2)(\sin{\left(x \right)})(\cos{\left(x \right)}) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%288%20x%20%2B%202%29%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%20" alt="LaTeX:  \displaystyle y = (8 x + 2)(\sin{\left(x \right)})(\cos{\left(x \right)}) " data-equation-content=" \displaystyle y = (8 x + 2)(\sin{\left(x \right)})(\cos{\left(x \right)}) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=8 x + 2 " src="/equation_images/%20%5Cdisplaystyle%20f%3D8%20x%20%2B%202%20" alt="LaTeX:  \displaystyle f=8 x + 2 " data-equation-content=" \displaystyle f=8 x + 2 " />  and  <img class="equation_image" title=" \displaystyle g=\sin{\left(x \right)} \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g=\sin{\left(x \right)} \cos{\left(x \right)} " data-equation-content=" \displaystyle g=\sin{\left(x \right)} \cos{\left(x \right)} " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=8 x + 2 \implies f'=8 " src="/equation_images/%20%5Cdisplaystyle%20f%3D8%20x%20%2B%202%20%5Cimplies%20f%27%3D8%20" alt="LaTeX:  \displaystyle f=8 x + 2 \implies f'=8 " data-equation-content=" \displaystyle f=8 x + 2 \implies f'=8 " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \sin{\left(x \right)} \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g = \sin{\left(x \right)} \cos{\left(x \right)} " data-equation-content=" \displaystyle g = \sin{\left(x \right)} \cos{\left(x \right)} " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cimplies%20f%27%3D%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)} " data-equation-content=" \displaystyle f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)} " />  and  <img class="equation_image" title=" \displaystyle g=\cos{\left(x \right)} \implies g'=- \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cimplies%20g%27%3D-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle g=\cos{\left(x \right)} \implies g'=- \sin{\left(x \right)} " data-equation-content=" \displaystyle g=\cos{\left(x \right)} \implies g'=- \sin{\left(x \right)} " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(\cos{\left(x \right)})(\cos{\left(x \right)})+(\sin{\left(x \right)})(- \sin{\left(x \right)}) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%20" alt="LaTeX:  \displaystyle g'=(\cos{\left(x \right)})(\cos{\left(x \right)})+(\sin{\left(x \right)})(- \sin{\left(x \right)}) " data-equation-content=" \displaystyle g'=(\cos{\left(x \right)})(\cos{\left(x \right)})+(\sin{\left(x \right)})(- \sin{\left(x \right)}) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(8 x + 2)(- \sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)})+(\sin{\left(x \right)} \cos{\left(x \right)})(8)=- \left(8 x + 2\right) \sin^{2}{\left(x \right)} + \left(8 x + 2\right) \cos^{2}{\left(x \right)} + 8 \sin{\left(x \right)} \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%288%20x%20%2B%202%29%28-%20%5Csin%5E%7B2%7D%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Ccos%5E%7B2%7D%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%288%29%3D-%20%5Cleft%288%20x%20%2B%202%5Cright%29%20%5Csin%5E%7B2%7D%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%288%20x%20%2B%202%5Cright%29%20%5Ccos%5E%7B2%7D%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%208%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f'=(8 x + 2)(- \sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)})+(\sin{\left(x \right)} \cos{\left(x \right)})(8)=- \left(8 x + 2\right) \sin^{2}{\left(x \right)} + \left(8 x + 2\right) \cos^{2}{\left(x \right)} + 8 \sin{\left(x \right)} \cos{\left(x \right)} " data-equation-content=" \displaystyle f'=(8 x + 2)(- \sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)})+(\sin{\left(x \right)} \cos{\left(x \right)})(8)=- \left(8 x + 2\right) \sin^{2}{\left(x \right)} + \left(8 x + 2\right) \cos^{2}{\left(x \right)} + 8 \sin{\left(x \right)} \cos{\left(x \right)} " /> </p> </p>