\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Find the derivative of \(\displaystyle y = (\sin{\left(x \right)})(- 4 x^{3} + 6 x^{2} + 6 x - 9)(\cos{\left(x \right)})\).
Identifying \(\displaystyle f=\sin{\left(x \right)}\) and \(\displaystyle g=\left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \right)}\) and using the product rule with \(\displaystyle f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)}\). This leaves g as \(\displaystyle g = \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \right)}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=- 4 x^{3} + 6 x^{2} + 6 x - 9 \implies f'=- 12 x^{2} + 12 x + 6\) and \(\displaystyle g=\cos{\left(x \right)} \implies g'=- \sin{\left(x \right)}\). Popping up a level gives \(\displaystyle g'=(\cos{\left(x \right)})(- 12 x^{2} + 12 x + 6)+(- 4 x^{3} + 6 x^{2} + 6 x - 9)(- \sin{\left(x \right)})\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(\sin{\left(x \right)})(\left(- 12 x^{2} + 12 x + 6\right) \cos{\left(x \right)} - \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \sin{\left(x \right)})+(\left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \right)})(\cos{\left(x \right)})=\left(- 12 x^{2} + 12 x + 6\right) \sin{\left(x \right)} \cos{\left(x \right)} - \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \sin^{2}{\left(x \right)} + \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos^{2}{\left(x \right)}\)
\begin{question}Find the derivative of $y = (\sin{\left(x \right)})(- 4 x^{3} + 6 x^{2} + 6 x - 9)(\cos{\left(x \right)})$.
\soln{9cm}{Identifying $f=\sin{\left(x \right)}$ and $g=\left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \right)}$ and using the product rule with $f=\sin{\left(x \right)} \implies f'=\cos{\left(x \right)}$. This leaves g as $g = \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \right)}$ which also requires the product rule. Pushing down in the new product rule $f=- 4 x^{3} + 6 x^{2} + 6 x - 9 \implies f'=- 12 x^{2} + 12 x + 6$ and $g=\cos{\left(x \right)} \implies g'=- \sin{\left(x \right)}$. Popping up a level gives $g'=(\cos{\left(x \right)})(- 12 x^{2} + 12 x + 6)+(- 4 x^{3} + 6 x^{2} + 6 x - 9)(- \sin{\left(x \right)})$Popping up again (Back to the original problem) gives $f'=(\sin{\left(x \right)})(\left(- 12 x^{2} + 12 x + 6\right) \cos{\left(x \right)} - \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \sin{\left(x \right)})+(\left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \right)})(\cos{\left(x \right)})=\left(- 12 x^{2} + 12 x + 6\right) \sin{\left(x \right)} \cos{\left(x \right)} - \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \sin^{2}{\left(x \right)} + \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos^{2}{\left(x \right)}$}
\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 = (\sin{\left(x \right)})(- 4 x^{3} + 6 x^{2} + 6 x - 9)(\cos{\left(x \right)}) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%204%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%20" alt="LaTeX: \displaystyle y = (\sin{\left(x \right)})(- 4 x^{3} + 6 x^{2} + 6 x - 9)(\cos{\left(x \right)}) " data-equation-content=" \displaystyle y = (\sin{\left(x \right)})(- 4 x^{3} + 6 x^{2} + 6 x - 9)(\cos{\left(x \right)}) " /> .</p> </p><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(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28-%204%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g=\left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \right)} " data-equation-content=" \displaystyle g=\left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \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(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28-%204%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g = \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \right)} " data-equation-content=" \displaystyle g = \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \right)} " /> which also requires the product rule. Pushing down in the new product rule <img class="equation_image" title=" \displaystyle f=- 4 x^{3} + 6 x^{2} + 6 x - 9 \implies f'=- 12 x^{2} + 12 x + 6 " src="/equation_images/%20%5Cdisplaystyle%20f%3D-%204%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%20%5Cimplies%20f%27%3D-%2012%20x%5E%7B2%7D%20%2B%2012%20x%20%2B%206%20" alt="LaTeX: \displaystyle f=- 4 x^{3} + 6 x^{2} + 6 x - 9 \implies f'=- 12 x^{2} + 12 x + 6 " data-equation-content=" \displaystyle f=- 4 x^{3} + 6 x^{2} + 6 x - 9 \implies f'=- 12 x^{2} + 12 x + 6 " /> and <img class="equation_image" title=" \displaystyle g=\cos{\left(x \right)} \implies g'=- \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cimplies%20g%27%3D-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g=\cos{\left(x \right)} \implies g'=- \sin{\left(x \right)} " data-equation-content=" \displaystyle g=\cos{\left(x \right)} \implies g'=- \sin{\left(x \right)} " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(\cos{\left(x \right)})(- 12 x^{2} + 12 x + 6)+(- 4 x^{3} + 6 x^{2} + 6 x - 9)(- \sin{\left(x \right)}) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%28-%2012%20x%5E%7B2%7D%20%2B%2012%20x%20%2B%206%29%2B%28-%204%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%29%28-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%20" alt="LaTeX: \displaystyle g'=(\cos{\left(x \right)})(- 12 x^{2} + 12 x + 6)+(- 4 x^{3} + 6 x^{2} + 6 x - 9)(- \sin{\left(x \right)}) " data-equation-content=" \displaystyle g'=(\cos{\left(x \right)})(- 12 x^{2} + 12 x + 6)+(- 4 x^{3} + 6 x^{2} + 6 x - 9)(- \sin{\left(x \right)}) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(\sin{\left(x \right)})(\left(- 12 x^{2} + 12 x + 6\right) \cos{\left(x \right)} - \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \sin{\left(x \right)})+(\left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \right)})(\cos{\left(x \right)})=\left(- 12 x^{2} + 12 x + 6\right) \sin{\left(x \right)} \cos{\left(x \right)} - \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \sin^{2}{\left(x \right)} + \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos^{2}{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%28%5Cleft%28-%2012%20x%5E%7B2%7D%20%2B%2012%20x%20%2B%206%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20-%20%5Cleft%28-%204%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Cleft%28-%204%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%3D%5Cleft%28-%2012%20x%5E%7B2%7D%20%2B%2012%20x%20%2B%206%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20-%20%5Cleft%28-%204%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%5Cright%29%20%5Csin%5E%7B2%7D%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%28-%204%20x%5E%7B3%7D%20%2B%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%5Cright%29%20%5Ccos%5E%7B2%7D%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle f'=(\sin{\left(x \right)})(\left(- 12 x^{2} + 12 x + 6\right) \cos{\left(x \right)} - \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \sin{\left(x \right)})+(\left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \right)})(\cos{\left(x \right)})=\left(- 12 x^{2} + 12 x + 6\right) \sin{\left(x \right)} \cos{\left(x \right)} - \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \sin^{2}{\left(x \right)} + \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos^{2}{\left(x \right)} " data-equation-content=" \displaystyle f'=(\sin{\left(x \right)})(\left(- 12 x^{2} + 12 x + 6\right) \cos{\left(x \right)} - \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \sin{\left(x \right)})+(\left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos{\left(x \right)})(\cos{\left(x \right)})=\left(- 12 x^{2} + 12 x + 6\right) \sin{\left(x \right)} \cos{\left(x \right)} - \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \sin^{2}{\left(x \right)} + \left(- 4 x^{3} + 6 x^{2} + 6 x - 9\right) \cos^{2}{\left(x \right)} " /> </p> </p>