Generate Math Exams and Quizzes in LaTeX and pdf formats

Identifying \(\displaystyle f=\cos{\left(x \right)}\) and \(\displaystyle g=\left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x}\) and using the product rule with \(\displaystyle f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)}\). This leaves g as \(\displaystyle g = \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=e^{x} \implies f'=e^{x}\) and \(\displaystyle g=3 x^{3} + 2 x^{2} - x - 8 \implies g'=9 x^{2} + 4 x - 1\). Popping up a level gives \(\displaystyle g'=(3 x^{3} + 2 x^{2} - x - 8)(e^{x})+(e^{x})(9 x^{2} + 4 x - 1)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(\cos{\left(x \right)})(\left(9 x^{2} + 4 x - 1\right) e^{x} + \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x})+(\left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x})(- \sin{\left(x \right)})=\left(9 x^{2} + 4 x - 1\right) e^{x} \cos{\left(x \right)} - \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} \sin{\left(x \right)} + \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} \cos{\left(x \right)}\)

Download \(\LaTeX\)

\begin{question}Find the derivative of $y = (\cos{\left(x \right)})(e^{x})(3 x^{3} + 2 x^{2} - x - 8)$.
    \soln{9cm}{Identifying $f=\cos{\left(x \right)}$ and $g=\left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x}$ and using the product rule with $f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)}$. This leaves g as $g = \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x}$ which also requires the product rule. Pushing down in the new product rule $f=e^{x} \implies f'=e^{x}$ and $g=3 x^{3} + 2 x^{2} - x - 8 \implies g'=9 x^{2} + 4 x - 1$. Popping up a level gives $g'=(3 x^{3} + 2 x^{2} - x - 8)(e^{x})+(e^{x})(9 x^{2} + 4 x - 1)$Popping up again (Back to the original problem) gives $f'=(\cos{\left(x \right)})(\left(9 x^{2} + 4 x - 1\right) e^{x} + \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x})+(\left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x})(- \sin{\left(x \right)})=\left(9 x^{2} + 4 x - 1\right) e^{x} \cos{\left(x \right)} - \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} \sin{\left(x \right)} + \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} \cos{\left(x \right)}$}

\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 = (\cos{\left(x \right)})(e^{x})(3 x^{3} + 2 x^{2} - x - 8) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%28e%5E%7Bx%7D%29%283%20x%5E%7B3%7D%20%2B%202%20x%5E%7B2%7D%20-%20x%20-%208%29%20" alt="LaTeX:  \displaystyle y = (\cos{\left(x \right)})(e^{x})(3 x^{3} + 2 x^{2} - x - 8) " data-equation-content=" \displaystyle y = (\cos{\left(x \right)})(e^{x})(3 x^{3} + 2 x^{2} - x - 8) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=\cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f=\cos{\left(x \right)} " data-equation-content=" \displaystyle f=\cos{\left(x \right)} " />  and  <img class="equation_image" title=" \displaystyle g=\left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%283%20x%5E%7B3%7D%20%2B%202%20x%5E%7B2%7D%20-%20x%20-%208%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g=\left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} " data-equation-content=" \displaystyle g=\left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cimplies%20f%27%3D-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)} " data-equation-content=" \displaystyle f=\cos{\left(x \right)} \implies f'=- \sin{\left(x \right)} " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%283%20x%5E%7B3%7D%20%2B%202%20x%5E%7B2%7D%20-%20x%20-%208%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g = \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} " data-equation-content=" \displaystyle g = \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=e^{x} \implies f'=e^{x} " src="/equation_images/%20%5Cdisplaystyle%20f%3De%5E%7Bx%7D%20%5Cimplies%20f%27%3De%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle f=e^{x} \implies f'=e^{x} " data-equation-content=" \displaystyle f=e^{x} \implies f'=e^{x} " />  and  <img class="equation_image" title=" \displaystyle g=3 x^{3} + 2 x^{2} - x - 8 \implies g'=9 x^{2} + 4 x - 1 " src="/equation_images/%20%5Cdisplaystyle%20g%3D3%20x%5E%7B3%7D%20%2B%202%20x%5E%7B2%7D%20-%20x%20-%208%20%5Cimplies%20g%27%3D9%20x%5E%7B2%7D%20%2B%204%20x%20-%201%20" alt="LaTeX:  \displaystyle g=3 x^{3} + 2 x^{2} - x - 8 \implies g'=9 x^{2} + 4 x - 1 " data-equation-content=" \displaystyle g=3 x^{3} + 2 x^{2} - x - 8 \implies g'=9 x^{2} + 4 x - 1 " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(3 x^{3} + 2 x^{2} - x - 8)(e^{x})+(e^{x})(9 x^{2} + 4 x - 1) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%283%20x%5E%7B3%7D%20%2B%202%20x%5E%7B2%7D%20-%20x%20-%208%29%28e%5E%7Bx%7D%29%2B%28e%5E%7Bx%7D%29%289%20x%5E%7B2%7D%20%2B%204%20x%20-%201%29%20" alt="LaTeX:  \displaystyle g'=(3 x^{3} + 2 x^{2} - x - 8)(e^{x})+(e^{x})(9 x^{2} + 4 x - 1) " data-equation-content=" \displaystyle g'=(3 x^{3} + 2 x^{2} - x - 8)(e^{x})+(e^{x})(9 x^{2} + 4 x - 1) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(\cos{\left(x \right)})(\left(9 x^{2} + 4 x - 1\right) e^{x} + \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x})+(\left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x})(- \sin{\left(x \right)})=\left(9 x^{2} + 4 x - 1\right) e^{x} \cos{\left(x \right)} - \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} \sin{\left(x \right)} + \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%28%5Cleft%289%20x%5E%7B2%7D%20%2B%204%20x%20-%201%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%283%20x%5E%7B3%7D%20%2B%202%20x%5E%7B2%7D%20-%20x%20-%208%5Cright%29%20e%5E%7Bx%7D%29%2B%28%5Cleft%283%20x%5E%7B3%7D%20%2B%202%20x%5E%7B2%7D%20-%20x%20-%208%5Cright%29%20e%5E%7Bx%7D%29%28-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%3D%5Cleft%289%20x%5E%7B2%7D%20%2B%204%20x%20-%201%5Cright%29%20e%5E%7Bx%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20-%20%5Cleft%283%20x%5E%7B3%7D%20%2B%202%20x%5E%7B2%7D%20-%20x%20-%208%5Cright%29%20e%5E%7Bx%7D%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%283%20x%5E%7B3%7D%20%2B%202%20x%5E%7B2%7D%20-%20x%20-%208%5Cright%29%20e%5E%7Bx%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f'=(\cos{\left(x \right)})(\left(9 x^{2} + 4 x - 1\right) e^{x} + \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x})+(\left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x})(- \sin{\left(x \right)})=\left(9 x^{2} + 4 x - 1\right) e^{x} \cos{\left(x \right)} - \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} \sin{\left(x \right)} + \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} \cos{\left(x \right)} " data-equation-content=" \displaystyle f'=(\cos{\left(x \right)})(\left(9 x^{2} + 4 x - 1\right) e^{x} + \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x})+(\left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x})(- \sin{\left(x \right)})=\left(9 x^{2} + 4 x - 1\right) e^{x} \cos{\left(x \right)} - \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} \sin{\left(x \right)} + \left(3 x^{3} + 2 x^{2} - x - 8\right) e^{x} \cos{\left(x \right)} " /> </p> </p>