\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Find the derivative of \(\displaystyle y = (5 x^{3} + 3 x^{2} + 5 x + 9)(4 x^{3} - 4 x^{2} - 7 x - 9)(\sin{\left(x \right)})\).
Identifying \(\displaystyle f=5 x^{3} + 3 x^{2} + 5 x + 9\) and \(\displaystyle g=\left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)}\) and using the product rule with \(\displaystyle f=5 x^{3} + 3 x^{2} + 5 x + 9 \implies f'=15 x^{2} + 6 x + 5\). This leaves g as \(\displaystyle g = \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)}\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=4 x^{3} - 4 x^{2} - 7 x - 9 \implies f'=12 x^{2} - 8 x - 7\) and \(\displaystyle g=\sin{\left(x \right)} \implies g'=\cos{\left(x \right)}\). Popping up a level gives \(\displaystyle g'=(\sin{\left(x \right)})(12 x^{2} - 8 x - 7)+(4 x^{3} - 4 x^{2} - 7 x - 9)(\cos{\left(x \right)})\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(5 x^{3} + 3 x^{2} + 5 x + 9)(\left(12 x^{2} - 8 x - 7\right) \sin{\left(x \right)} + \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \cos{\left(x \right)})+(\left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)})(15 x^{2} + 6 x + 5)=\left(12 x^{2} - 8 x - 7\right) \left(5 x^{3} + 3 x^{2} + 5 x + 9\right) \sin{\left(x \right)} + \left(15 x^{2} + 6 x + 5\right) \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)} + \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \left(5 x^{3} + 3 x^{2} + 5 x + 9\right) \cos{\left(x \right)}\)
\begin{question}Find the derivative of $y = (5 x^{3} + 3 x^{2} + 5 x + 9)(4 x^{3} - 4 x^{2} - 7 x - 9)(\sin{\left(x \right)})$.
\soln{9cm}{Identifying $f=5 x^{3} + 3 x^{2} + 5 x + 9$ and $g=\left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)}$ and using the product rule with $f=5 x^{3} + 3 x^{2} + 5 x + 9 \implies f'=15 x^{2} + 6 x + 5$. This leaves g as $g = \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)}$ which also requires the product rule. Pushing down in the new product rule $f=4 x^{3} - 4 x^{2} - 7 x - 9 \implies f'=12 x^{2} - 8 x - 7$ and $g=\sin{\left(x \right)} \implies g'=\cos{\left(x \right)}$. Popping up a level gives $g'=(\sin{\left(x \right)})(12 x^{2} - 8 x - 7)+(4 x^{3} - 4 x^{2} - 7 x - 9)(\cos{\left(x \right)})$Popping up again (Back to the original problem) gives $f'=(5 x^{3} + 3 x^{2} + 5 x + 9)(\left(12 x^{2} - 8 x - 7\right) \sin{\left(x \right)} + \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \cos{\left(x \right)})+(\left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)})(15 x^{2} + 6 x + 5)=\left(12 x^{2} - 8 x - 7\right) \left(5 x^{3} + 3 x^{2} + 5 x + 9\right) \sin{\left(x \right)} + \left(15 x^{2} + 6 x + 5\right) \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)} + \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \left(5 x^{3} + 3 x^{2} + 5 x + 9\right) \cos{\left(x \right)}$}
\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 = (5 x^{3} + 3 x^{2} + 5 x + 9)(4 x^{3} - 4 x^{2} - 7 x - 9)(\sin{\left(x \right)}) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%285%20x%5E%7B3%7D%20%2B%203%20x%5E%7B2%7D%20%2B%205%20x%20%2B%209%29%284%20x%5E%7B3%7D%20-%204%20x%5E%7B2%7D%20-%207%20x%20-%209%29%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%20" alt="LaTeX: \displaystyle y = (5 x^{3} + 3 x^{2} + 5 x + 9)(4 x^{3} - 4 x^{2} - 7 x - 9)(\sin{\left(x \right)}) " data-equation-content=" \displaystyle y = (5 x^{3} + 3 x^{2} + 5 x + 9)(4 x^{3} - 4 x^{2} - 7 x - 9)(\sin{\left(x \right)}) " /> .</p> </p><p> <p>Identifying <img class="equation_image" title=" \displaystyle f=5 x^{3} + 3 x^{2} + 5 x + 9 " src="/equation_images/%20%5Cdisplaystyle%20f%3D5%20x%5E%7B3%7D%20%2B%203%20x%5E%7B2%7D%20%2B%205%20x%20%2B%209%20" alt="LaTeX: \displaystyle f=5 x^{3} + 3 x^{2} + 5 x + 9 " data-equation-content=" \displaystyle f=5 x^{3} + 3 x^{2} + 5 x + 9 " /> and <img class="equation_image" title=" \displaystyle g=\left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%284%20x%5E%7B3%7D%20-%204%20x%5E%7B2%7D%20-%207%20x%20-%209%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g=\left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle g=\left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)} " /> and using the product rule with <img class="equation_image" title=" \displaystyle f=5 x^{3} + 3 x^{2} + 5 x + 9 \implies f'=15 x^{2} + 6 x + 5 " src="/equation_images/%20%5Cdisplaystyle%20f%3D5%20x%5E%7B3%7D%20%2B%203%20x%5E%7B2%7D%20%2B%205%20x%20%2B%209%20%5Cimplies%20f%27%3D15%20x%5E%7B2%7D%20%2B%206%20x%20%2B%205%20" alt="LaTeX: \displaystyle f=5 x^{3} + 3 x^{2} + 5 x + 9 \implies f'=15 x^{2} + 6 x + 5 " data-equation-content=" \displaystyle f=5 x^{3} + 3 x^{2} + 5 x + 9 \implies f'=15 x^{2} + 6 x + 5 " /> . This leaves g as <img class="equation_image" title=" \displaystyle g = \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%284%20x%5E%7B3%7D%20-%204%20x%5E%7B2%7D%20-%207%20x%20-%209%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g = \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle g = \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\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} - 4 x^{2} - 7 x - 9 \implies f'=12 x^{2} - 8 x - 7 " src="/equation_images/%20%5Cdisplaystyle%20f%3D4%20x%5E%7B3%7D%20-%204%20x%5E%7B2%7D%20-%207%20x%20-%209%20%5Cimplies%20f%27%3D12%20x%5E%7B2%7D%20-%208%20x%20-%207%20" alt="LaTeX: \displaystyle f=4 x^{3} - 4 x^{2} - 7 x - 9 \implies f'=12 x^{2} - 8 x - 7 " data-equation-content=" \displaystyle f=4 x^{3} - 4 x^{2} - 7 x - 9 \implies f'=12 x^{2} - 8 x - 7 " /> and <img class="equation_image" title=" \displaystyle g=\sin{\left(x \right)} \implies g'=\cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cimplies%20g%27%3D%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g=\sin{\left(x \right)} \implies g'=\cos{\left(x \right)} " data-equation-content=" \displaystyle g=\sin{\left(x \right)} \implies g'=\cos{\left(x \right)} " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(\sin{\left(x \right)})(12 x^{2} - 8 x - 7)+(4 x^{3} - 4 x^{2} - 7 x - 9)(\cos{\left(x \right)}) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%2812%20x%5E%7B2%7D%20-%208%20x%20-%207%29%2B%284%20x%5E%7B3%7D%20-%204%20x%5E%7B2%7D%20-%207%20x%20-%209%29%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%20" alt="LaTeX: \displaystyle g'=(\sin{\left(x \right)})(12 x^{2} - 8 x - 7)+(4 x^{3} - 4 x^{2} - 7 x - 9)(\cos{\left(x \right)}) " data-equation-content=" \displaystyle g'=(\sin{\left(x \right)})(12 x^{2} - 8 x - 7)+(4 x^{3} - 4 x^{2} - 7 x - 9)(\cos{\left(x \right)}) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(5 x^{3} + 3 x^{2} + 5 x + 9)(\left(12 x^{2} - 8 x - 7\right) \sin{\left(x \right)} + \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \cos{\left(x \right)})+(\left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)})(15 x^{2} + 6 x + 5)=\left(12 x^{2} - 8 x - 7\right) \left(5 x^{3} + 3 x^{2} + 5 x + 9\right) \sin{\left(x \right)} + \left(15 x^{2} + 6 x + 5\right) \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)} + \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \left(5 x^{3} + 3 x^{2} + 5 x + 9\right) \cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%285%20x%5E%7B3%7D%20%2B%203%20x%5E%7B2%7D%20%2B%205%20x%20%2B%209%29%28%5Cleft%2812%20x%5E%7B2%7D%20-%208%20x%20-%207%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%284%20x%5E%7B3%7D%20-%204%20x%5E%7B2%7D%20-%207%20x%20-%209%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%29%2B%28%5Cleft%284%20x%5E%7B3%7D%20-%204%20x%5E%7B2%7D%20-%207%20x%20-%209%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%29%2815%20x%5E%7B2%7D%20%2B%206%20x%20%2B%205%29%3D%5Cleft%2812%20x%5E%7B2%7D%20-%208%20x%20-%207%5Cright%29%20%5Cleft%285%20x%5E%7B3%7D%20%2B%203%20x%5E%7B2%7D%20%2B%205%20x%20%2B%209%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%2815%20x%5E%7B2%7D%20%2B%206%20x%20%2B%205%5Cright%29%20%5Cleft%284%20x%5E%7B3%7D%20-%204%20x%5E%7B2%7D%20-%207%20x%20-%209%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cleft%284%20x%5E%7B3%7D%20-%204%20x%5E%7B2%7D%20-%207%20x%20-%209%5Cright%29%20%5Cleft%285%20x%5E%7B3%7D%20%2B%203%20x%5E%7B2%7D%20%2B%205%20x%20%2B%209%5Cright%29%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle f'=(5 x^{3} + 3 x^{2} + 5 x + 9)(\left(12 x^{2} - 8 x - 7\right) \sin{\left(x \right)} + \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \cos{\left(x \right)})+(\left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)})(15 x^{2} + 6 x + 5)=\left(12 x^{2} - 8 x - 7\right) \left(5 x^{3} + 3 x^{2} + 5 x + 9\right) \sin{\left(x \right)} + \left(15 x^{2} + 6 x + 5\right) \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)} + \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \left(5 x^{3} + 3 x^{2} + 5 x + 9\right) \cos{\left(x \right)} " data-equation-content=" \displaystyle f'=(5 x^{3} + 3 x^{2} + 5 x + 9)(\left(12 x^{2} - 8 x - 7\right) \sin{\left(x \right)} + \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \cos{\left(x \right)})+(\left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)})(15 x^{2} + 6 x + 5)=\left(12 x^{2} - 8 x - 7\right) \left(5 x^{3} + 3 x^{2} + 5 x + 9\right) \sin{\left(x \right)} + \left(15 x^{2} + 6 x + 5\right) \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \sin{\left(x \right)} + \left(4 x^{3} - 4 x^{2} - 7 x - 9\right) \left(5 x^{3} + 3 x^{2} + 5 x + 9\right) \cos{\left(x \right)} " /> </p> </p>