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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y = e^{- x} \cos{\left(x \right)}\).
Using the quotient rule with \(\displaystyle f = \cos{\left(x \right)}\), \(\displaystyle f' = - \sin{\left(x \right)}\), \(\displaystyle g = e^{x}\), and \(\displaystyle g'= e^{x}\) gives:
\begin{equation*} \frac{ (e^{x})(- \sin{\left(x \right)}) - (\cos{\left(x \right)})(e^{x})}{(e^{x})^2} = - \left(\sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{- x} \end{equation*}
\begin{question}Find the derivative of $y = e^{- x} \cos{\left(x \right)}$. \soln{9cm}{Using the quotient rule with $f = \cos{\left(x \right)}$, $f' = - \sin{\left(x \right)}$, $g = e^{x}$, and $g'= e^{x}$ gives:\newline \begin{equation*} \frac{ (e^{x})(- \sin{\left(x \right)}) - (\cos{\left(x \right)})(e^{x})}{(e^{x})^2} = - \left(\sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{- x} \end{equation*}} \end{question}
\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}
<p> <p>Find the derivative of <img class="equation_image" title=" \displaystyle y = e^{- x} \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20e%5E%7B-%20x%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle y = e^{- x} \cos{\left(x \right)} " data-equation-content=" \displaystyle y = e^{- x} \cos{\left(x \right)} " /> . </p> </p>
<p> <p>Using the quotient rule with <img class="equation_image" title=" \displaystyle f = \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%20%3D%20%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)} " /> , <img class="equation_image" title=" \displaystyle f' = - \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%20%3D%20-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle f' = - \sin{\left(x \right)} " data-equation-content=" \displaystyle f' = - \sin{\left(x \right)} " /> , <img class="equation_image" title=" \displaystyle g = e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20e%5E%7Bx%7D%20" alt="LaTeX: \displaystyle g = e^{x} " data-equation-content=" \displaystyle g = e^{x} " /> , and <img class="equation_image" title=" \displaystyle g'= e^{x} " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%20e%5E%7Bx%7D%20" alt="LaTeX: \displaystyle g'= e^{x} " data-equation-content=" \displaystyle g'= e^{x} " /> gives:<br>
<img class="equation_image" title=" \frac{ (e^{x})(- \sin{\left(x \right)}) - (\cos{\left(x \right)})(e^{x})}{(e^{x})^2} = - \left(\sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{- x} " src="/equation_images/%20%20%5Cfrac%7B%20%28e%5E%7Bx%7D%29%28-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%20-%20%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%28e%5E%7Bx%7D%29%7D%7B%28e%5E%7Bx%7D%29%5E2%7D%20%3D%20-%20%5Cleft%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%5Cright%29%20e%5E%7B-%20x%7D%20%20" alt="LaTeX: \frac{ (e^{x})(- \sin{\left(x \right)}) - (\cos{\left(x \right)})(e^{x})}{(e^{x})^2} = - \left(\sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{- x} " data-equation-content=" \frac{ (e^{x})(- \sin{\left(x \right)}) - (\cos{\left(x \right)})(e^{x})}{(e^{x})^2} = - \left(\sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{- x} " /> </p> </p>