\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus

Please login to create an exam or a quiz.

Calculus
Derivatives
New Random

Find the derivative of \(\displaystyle y = (\sin{\left(x \right)})(2 x + 8)(6 x + 1)\).


Identifying \(\displaystyle f=\sin{\left(x \right)}\) and \(\displaystyle g=\left(2 x + 8\right) \left(6 x + 1\right)\) and using the product rule with \(\displaystyle f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)}\). This leaves g as \(\displaystyle g = \left(2 x + 8\right) \left(6 x + 1\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=2 x + 8 \implies f'=2\) and \(\displaystyle g=6 x + 1 \implies g'=6\). Popping up a level gives \(\displaystyle g'=(6 x + 1)(2)+(2 x + 8)(6)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(\sin{\left(x \right)})(24 x + 50)+(\left(2 x + 8\right) \left(6 x + 1\right))(\cos{\left(x \right)})=\left(2 x + 8\right) \left(6 x + 1\right) \cos{\left(x \right)} + \left(12 x + 2\right) \sin{\left(x \right)} + \left(12 x + 48\right) \sin{\left(x \right)}\)

Download \(\LaTeX\)

\begin{question}Find the derivative of $y = (\sin{\left(x \right)})(2 x + 8)(6 x + 1)$.
    \soln{9cm}{Identifying $f=\sin{\left(x \right)}$ and $g=\left(2 x + 8\right) \left(6 x + 1\right)$ and using the product rule with $f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)}$. This leaves g as $g = \left(2 x + 8\right) \left(6 x + 1\right)$ which also requires the product rule. Pushing down in the new product rule $f=2 x + 8 \implies f'=2$ and $g=6 x + 1 \implies g'=6$. Popping up a level gives $g'=(6 x + 1)(2)+(2 x + 8)(6)$Popping up again (Back to the original problem) gives $f'=(\sin{\left(x \right)})(24 x + 50)+(\left(2 x + 8\right) \left(6 x + 1\right))(\cos{\left(x \right)})=\left(2 x + 8\right) \left(6 x + 1\right) \cos{\left(x \right)} + \left(12 x + 2\right) \sin{\left(x \right)} + \left(12 x + 48\right) \sin{\left(x \right)}$}

\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 = (\sin{\left(x \right)})(2 x + 8)(6 x + 1) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%282%20x%20%2B%208%29%286%20x%20%2B%201%29%20" alt="LaTeX:  \displaystyle y = (\sin{\left(x \right)})(2 x + 8)(6 x + 1) " data-equation-content=" \displaystyle y = (\sin{\left(x \right)})(2 x + 8)(6 x + 1) " /> .</p> </p>
HTML for Canvas
<p> <p>Identifying  <img class="equation_image" title=" \displaystyle f=\sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%3D%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)} " />  and  <img class="equation_image" title=" \displaystyle g=\left(2 x + 8\right) \left(6 x + 1\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%282%20x%20%2B%208%5Cright%29%20%5Cleft%286%20x%20%2B%201%5Cright%29%20" alt="LaTeX:  \displaystyle g=\left(2 x + 8\right) \left(6 x + 1\right) " data-equation-content=" \displaystyle g=\left(2 x + 8\right) \left(6 x + 1\right) " />  and using the product rule with  <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)} " /> . This leaves g as  <img class="equation_image" title=" \displaystyle g = \left(2 x + 8\right) \left(6 x + 1\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%282%20x%20%2B%208%5Cright%29%20%5Cleft%286%20x%20%2B%201%5Cright%29%20" alt="LaTeX:  \displaystyle g = \left(2 x + 8\right) \left(6 x + 1\right) " data-equation-content=" \displaystyle g = \left(2 x + 8\right) \left(6 x + 1\right) " />  which also requires the product rule. Pushing down in the new product rule  <img class="equation_image" title=" \displaystyle f=2 x + 8 \implies f'=2 " src="/equation_images/%20%5Cdisplaystyle%20f%3D2%20x%20%2B%208%20%5Cimplies%20f%27%3D2%20" alt="LaTeX:  \displaystyle f=2 x + 8 \implies f'=2 " data-equation-content=" \displaystyle f=2 x + 8 \implies f'=2 " />  and  <img class="equation_image" title=" \displaystyle g=6 x + 1 \implies g'=6 " src="/equation_images/%20%5Cdisplaystyle%20g%3D6%20x%20%2B%201%20%5Cimplies%20g%27%3D6%20" alt="LaTeX:  \displaystyle g=6 x + 1 \implies g'=6 " data-equation-content=" \displaystyle g=6 x + 1 \implies g'=6 " /> . Popping up a level gives  <img class="equation_image" title=" \displaystyle g'=(6 x + 1)(2)+(2 x + 8)(6) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%286%20x%20%2B%201%29%282%29%2B%282%20x%20%2B%208%29%286%29%20" alt="LaTeX:  \displaystyle g'=(6 x + 1)(2)+(2 x + 8)(6) " data-equation-content=" \displaystyle g'=(6 x + 1)(2)+(2 x + 8)(6) " /> Popping up again (Back to the original problem) gives  <img class="equation_image" title=" \displaystyle f'=(\sin{\left(x \right)})(24 x + 50)+(\left(2 x + 8\right) \left(6 x + 1\right))(\cos{\left(x \right)})=\left(2 x + 8\right) \left(6 x + 1\right) \cos{\left(x \right)} + \left(12 x + 2\right) \sin{\left(x \right)} + \left(12 x + 48\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%2824%20x%20%2B%2050%29%2B%28%5Cleft%282%20x%20%2B%208%5Cright%29%20%5Cleft%286%20x%20%2B%201%5Cright%29%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%3D%5Cleft%282%20x%20%2B%208%5Cright%29%20%5Cleft%286%20x%20%2B%201%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2812%20x%20%2B%202%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2812%20x%20%2B%2048%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle f'=(\sin{\left(x \right)})(24 x + 50)+(\left(2 x + 8\right) \left(6 x + 1\right))(\cos{\left(x \right)})=\left(2 x + 8\right) \left(6 x + 1\right) \cos{\left(x \right)} + \left(12 x + 2\right) \sin{\left(x \right)} + \left(12 x + 48\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle f'=(\sin{\left(x \right)})(24 x + 50)+(\left(2 x + 8\right) \left(6 x + 1\right))(\cos{\left(x \right)})=\left(2 x + 8\right) \left(6 x + 1\right) \cos{\left(x \right)} + \left(12 x + 2\right) \sin{\left(x \right)} + \left(12 x + 48\right) \sin{\left(x \right)} " /> </p> </p>