Generate Math Exams and Quizzes in LaTeX and pdf formats

Identifying \(\displaystyle f=\log{\left(x \right)}\) and \(\displaystyle g=\left(x^{2} + 5 x + 9\right) e^{x}\) and using the product rule with \(\displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x}\). This leaves g as \(\displaystyle g = \left(x^{2} + 5 x + 9\right) e^{x}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=e^{x} \implies f'=e^{x}\) and \(\displaystyle g=x^{2} + 5 x + 9 \implies g'=2 x + 5\). Popping up a level gives \(\displaystyle g'=(x^{2} + 5 x + 9)(e^{x})+(e^{x})(2 x + 5)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(\log{\left(x \right)})(\left(2 x + 5\right) e^{x} + \left(x^{2} + 5 x + 9\right) e^{x})+(\left(x^{2} + 5 x + 9\right) e^{x})(\frac{1}{x})=\left(2 x + 5\right) e^{x} \log{\left(x \right)} + \left(x^{2} + 5 x + 9\right) e^{x} \log{\left(x \right)} + \frac{\left(x^{2} + 5 x + 9\right) e^{x}}{x}\)

Download \(\LaTeX\)

\begin{question}Find the derivative of $y = (\log{\left(x \right)})(e^{x})(x^{2} + 5 x + 9)$.
    \soln{9cm}{Identifying $f=\log{\left(x \right)}$ and $g=\left(x^{2} + 5 x + 9\right) e^{x}$ and using the product rule with $f=\log{\left(x \right)} \implies f'=\frac{1}{x}$. This leaves g as $g = \left(x^{2} + 5 x + 9\right) e^{x}$ which also requires the product rule. Pushing down in the new product rule $f=e^{x} \implies f'=e^{x}$ and $g=x^{2} + 5 x + 9 \implies g'=2 x + 5$. Popping up a level gives $g'=(x^{2} + 5 x + 9)(e^{x})+(e^{x})(2 x + 5)$Popping up again (Back to the original problem) gives $f'=(\log{\left(x \right)})(\left(2 x + 5\right) e^{x} + \left(x^{2} + 5 x + 9\right) e^{x})+(\left(x^{2} + 5 x + 9\right) e^{x})(\frac{1}{x})=\left(2 x + 5\right) e^{x} \log{\left(x \right)} + \left(x^{2} + 5 x + 9\right) e^{x} \log{\left(x \right)} + \frac{\left(x^{2} + 5 x + 9\right) e^{x}}{x}$}

\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 = (\log{\left(x \right)})(e^{x})(x^{2} + 5 x + 9) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%28e%5E%7Bx%7D%29%28x%5E%7B2%7D%20%2B%205%20x%20%2B%209%29%20" alt="LaTeX:  \displaystyle y = (\log{\left(x \right)})(e^{x})(x^{2} + 5 x + 9) " data-equation-content=" \displaystyle y = (\log{\left(x \right)})(e^{x})(x^{2} + 5 x + 9) " /> .</p> </p>

HTML for Canvas

<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=\log{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f=\log{\left(x \right)} " data-equation-content=" \displaystyle f=\log{\left(x \right)} " />  and  <img class="equation_image" title=" \displaystyle g=\left(x^{2} + 5 x + 9\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28x%5E%7B2%7D%20%2B%205%20x%20%2B%209%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g=\left(x^{2} + 5 x + 9\right) e^{x} " data-equation-content=" \displaystyle g=\left(x^{2} + 5 x + 9\right) e^{x} " />  and using the product rule with  <img class="equation_image" title=" \displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cimplies%20f%27%3D%5Cfrac%7B1%7D%7Bx%7D%20" alt="LaTeX:  \displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x} " data-equation-content=" \displaystyle f=\log{\left(x \right)} \implies f'=\frac{1}{x} " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(x^{2} + 5 x + 9\right) e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28x%5E%7B2%7D%20%2B%205%20x%20%2B%209%5Cright%29%20e%5E%7Bx%7D%20" alt="LaTeX:  \displaystyle g = \left(x^{2} + 5 x + 9\right) e^{x} " data-equation-content=" \displaystyle g = \left(x^{2} + 5 x + 9\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=x^{2} + 5 x + 9 \implies g'=2 x + 5 " src="/equation_images/%20%5Cdisplaystyle%20g%3Dx%5E%7B2%7D%20%2B%205%20x%20%2B%209%20%5Cimplies%20g%27%3D2%20x%20%2B%205%20" alt="LaTeX:  \displaystyle g=x^{2} + 5 x + 9 \implies g'=2 x + 5 " data-equation-content=" \displaystyle g=x^{2} + 5 x + 9 \implies g'=2 x + 5 " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(x^{2} + 5 x + 9)(e^{x})+(e^{x})(2 x + 5) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28x%5E%7B2%7D%20%2B%205%20x%20%2B%209%29%28e%5E%7Bx%7D%29%2B%28e%5E%7Bx%7D%29%282%20x%20%2B%205%29%20" alt="LaTeX:  \displaystyle g'=(x^{2} + 5 x + 9)(e^{x})+(e^{x})(2 x + 5) " data-equation-content=" \displaystyle g'=(x^{2} + 5 x + 9)(e^{x})+(e^{x})(2 x + 5) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(\log{\left(x \right)})(\left(2 x + 5\right) e^{x} + \left(x^{2} + 5 x + 9\right) e^{x})+(\left(x^{2} + 5 x + 9\right) e^{x})(\frac{1}{x})=\left(2 x + 5\right) e^{x} \log{\left(x \right)} + \left(x^{2} + 5 x + 9\right) e^{x} \log{\left(x \right)} + \frac{\left(x^{2} + 5 x + 9\right) e^{x}}{x} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%29%28%5Cleft%282%20x%20%2B%205%5Cright%29%20e%5E%7Bx%7D%20%2B%20%5Cleft%28x%5E%7B2%7D%20%2B%205%20x%20%2B%209%5Cright%29%20e%5E%7Bx%7D%29%2B%28%5Cleft%28x%5E%7B2%7D%20%2B%205%20x%20%2B%209%5Cright%29%20e%5E%7Bx%7D%29%28%5Cfrac%7B1%7D%7Bx%7D%29%3D%5Cleft%282%20x%20%2B%205%5Cright%29%20e%5E%7Bx%7D%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%28x%5E%7B2%7D%20%2B%205%20x%20%2B%209%5Cright%29%20e%5E%7Bx%7D%20%5Clog%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B%5Cleft%28x%5E%7B2%7D%20%2B%205%20x%20%2B%209%5Cright%29%20e%5E%7Bx%7D%7D%7Bx%7D%20" alt="LaTeX:  \displaystyle f'=(\log{\left(x \right)})(\left(2 x + 5\right) e^{x} + \left(x^{2} + 5 x + 9\right) e^{x})+(\left(x^{2} + 5 x + 9\right) e^{x})(\frac{1}{x})=\left(2 x + 5\right) e^{x} \log{\left(x \right)} + \left(x^{2} + 5 x + 9\right) e^{x} \log{\left(x \right)} + \frac{\left(x^{2} + 5 x + 9\right) e^{x}}{x} " data-equation-content=" \displaystyle f'=(\log{\left(x \right)})(\left(2 x + 5\right) e^{x} + \left(x^{2} + 5 x + 9\right) e^{x})+(\left(x^{2} + 5 x + 9\right) e^{x})(\frac{1}{x})=\left(2 x + 5\right) e^{x} \log{\left(x \right)} + \left(x^{2} + 5 x + 9\right) e^{x} \log{\left(x \right)} + \frac{\left(x^{2} + 5 x + 9\right) e^{x}}{x} " /> </p> </p>