Generate Math Exams and Quizzes in LaTeX and pdf formats

Decomposing the function gives $f (u) = e^{u}$ , $u = \cos (v)$ , and $v = x^{5} .$ Using the chain rule $f^{'} (x) = \frac{d f}{d u} \frac{d u}{d v} \frac{d v}{d x}$ . $f^{'} (x) = (e^{u}) (- \sin (v)) (5 x^{4}) = - 5 x^{4} e^{u} \sin (v)$ . Substituting back in $u$ and $v$ gives $f^{'} (x) = - 5 x^{4} e^{\cos (v)} \sin (v) = - 5 x^{4} e^{\cos (x^{5})} \sin (x^{5})$ .

Download

L A T E X

\begin{question}Find the derivative of  $f(x) = e^{\cos{\left(x^{5} \right)}}$. 
    \soln{9cm}{Decomposing the function gives $f(u) = e^{u}$, $u = \cos{\left(v \right)}$, and $ v = x^{5}.$ Using the chain rule $f'(x) = \frac{df}{du}\frac{du}{dv}\frac{dv}{dx}$. $f'(x) = (e^{u})(- \sin{\left(v \right)})(5 x^{4}) = - 5 x^{4} e^{u} \sin{\left(v \right)}$. Substituting back in $u$ and $v$ gives $f'(x) = - 5 x^{4} e^{\cos{\left(v \right)}} \sin{\left(v \right)} = - 5 x^{4} e^{\cos{\left(x^{5} \right)}} \sin{\left(x^{5} \right)}$. }

\end{question}

Download Question and Solution Environment

L A T E X

\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 f(x) = e^{\cos{\left(x^{5} \right)}} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20e%5E%7B%5Ccos%7B%5Cleft%28x%5E%7B5%7D%20%5Cright%29%7D%7D%20" alt="LaTeX:  \displaystyle f(x) = e^{\cos{\left(x^{5} \right)}} " data-equation-content=" \displaystyle f(x) = e^{\cos{\left(x^{5} \right)}} " /> . </p> </p>

HTML for Canvas

<p> <p>Decomposing the function gives  <img class="equation_image" title=" \displaystyle f(u) = e^{u} " src="/equation_images/%20%5Cdisplaystyle%20f%28u%29%20%3D%20e%5E%7Bu%7D%20" alt="LaTeX:  \displaystyle f(u) = e^{u} " data-equation-content=" \displaystyle f(u) = e^{u} " /> ,  <img class="equation_image" title=" \displaystyle u = \cos{\left(v \right)} " src="/equation_images/%20%5Cdisplaystyle%20u%20%3D%20%5Ccos%7B%5Cleft%28v%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle u = \cos{\left(v \right)} " data-equation-content=" \displaystyle u = \cos{\left(v \right)} " /> , and  <img class="equation_image" title=" \displaystyle  v = x^{5}. " src="/equation_images/%20%5Cdisplaystyle%20%20v%20%3D%20x%5E%7B5%7D.%20" alt="LaTeX:  \displaystyle  v = x^{5}. " data-equation-content=" \displaystyle  v = x^{5}. " />  Using the chain rule  <img class="equation_image" title=" \displaystyle f'(x) = \frac{df}{du}\frac{du}{dv}\frac{dv}{dx} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7Bdf%7D%7Bdu%7D%5Cfrac%7Bdu%7D%7Bdv%7D%5Cfrac%7Bdv%7D%7Bdx%7D%20" alt="LaTeX:  \displaystyle f'(x) = \frac{df}{du}\frac{du}{dv}\frac{dv}{dx} " data-equation-content=" \displaystyle f'(x) = \frac{df}{du}\frac{du}{dv}\frac{dv}{dx} " /> .  <img class="equation_image" title=" \displaystyle f'(x) = (e^{u})(- \sin{\left(v \right)})(5 x^{4}) = - 5 x^{4} e^{u} \sin{\left(v \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%28e%5E%7Bu%7D%29%28-%20%5Csin%7B%5Cleft%28v%20%5Cright%29%7D%29%285%20x%5E%7B4%7D%29%20%3D%20-%205%20x%5E%7B4%7D%20e%5E%7Bu%7D%20%5Csin%7B%5Cleft%28v%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f'(x) = (e^{u})(- \sin{\left(v \right)})(5 x^{4}) = - 5 x^{4} e^{u} \sin{\left(v \right)} " data-equation-content=" \displaystyle f'(x) = (e^{u})(- \sin{\left(v \right)})(5 x^{4}) = - 5 x^{4} e^{u} \sin{\left(v \right)} " /> . Substituting back in  <img class="equation_image" title=" \displaystyle u " src="/equation_images/%20%5Cdisplaystyle%20u%20" alt="LaTeX:  \displaystyle u " data-equation-content=" \displaystyle u " />  and  <img class="equation_image" title=" \displaystyle v " src="/equation_images/%20%5Cdisplaystyle%20v%20" alt="LaTeX:  \displaystyle v " data-equation-content=" \displaystyle v " />  gives  <img class="equation_image" title=" \displaystyle f'(x) = - 5 x^{4} e^{\cos{\left(v \right)}} \sin{\left(v \right)} = - 5 x^{4} e^{\cos{\left(x^{5} \right)}} \sin{\left(x^{5} \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20-%205%20x%5E%7B4%7D%20e%5E%7B%5Ccos%7B%5Cleft%28v%20%5Cright%29%7D%7D%20%5Csin%7B%5Cleft%28v%20%5Cright%29%7D%20%3D%20-%205%20x%5E%7B4%7D%20e%5E%7B%5Ccos%7B%5Cleft%28x%5E%7B5%7D%20%5Cright%29%7D%7D%20%5Csin%7B%5Cleft%28x%5E%7B5%7D%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f'(x) = - 5 x^{4} e^{\cos{\left(v \right)}} \sin{\left(v \right)} = - 5 x^{4} e^{\cos{\left(x^{5} \right)}} \sin{\left(x^{5} \right)} " data-equation-content=" \displaystyle f'(x) = - 5 x^{4} e^{\cos{\left(v \right)}} \sin{\left(v \right)} = - 5 x^{4} e^{\cos{\left(x^{5} \right)}} \sin{\left(x^{5} \right)} " /> . </p> </p>