\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Find the derivative of \(\displaystyle y = (x^{3} - 2 x^{2} - 5 x - 4)(5 x^{3} - 6 x^{2} + 6 x - 9)(- 5 x^{3} - x^{2} - 6 x + 7)\).
Identifying \(\displaystyle f=x^{3} - 2 x^{2} - 5 x - 4\) and \(\displaystyle g=\left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right)\) and using the product rule with \(\displaystyle f=x^{3} - 2 x^{2} - 5 x - 4 \implies f'=3 x^{2} - 4 x - 5\). This leaves g as \(\displaystyle g = \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right)\) which also requires the product rule. Pushing down in the new product rule \(\displaystyle f=5 x^{3} - 6 x^{2} + 6 x - 9 \implies f'=15 x^{2} - 12 x + 6\) and \(\displaystyle g=- 5 x^{3} - x^{2} - 6 x + 7 \implies g'=- 15 x^{2} - 2 x - 6\). Popping up a level gives \(\displaystyle g'=(- 5 x^{3} - x^{2} - 6 x + 7)(15 x^{2} - 12 x + 6)+(5 x^{3} - 6 x^{2} + 6 x - 9)(- 15 x^{2} - 2 x - 6)\)Popping up again (Back to the original problem) gives \(\displaystyle f'=(x^{3} - 2 x^{2} - 5 x - 4)(\left(- 15 x^{2} - 2 x - 6\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(15 x^{2} - 12 x + 6\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right))+(\left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right))(3 x^{2} - 4 x - 5)=\left(- 15 x^{2} - 2 x - 6\right) \left(x^{3} - 2 x^{2} - 5 x - 4\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(3 x^{2} - 4 x - 5\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(15 x^{2} - 12 x + 6\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(x^{3} - 2 x^{2} - 5 x - 4\right)\)
\begin{question}Find the derivative of $y = (x^{3} - 2 x^{2} - 5 x - 4)(5 x^{3} - 6 x^{2} + 6 x - 9)(- 5 x^{3} - x^{2} - 6 x + 7)$.
\soln{9cm}{Identifying $f=x^{3} - 2 x^{2} - 5 x - 4$ and $g=\left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right)$ and using the product rule with $f=x^{3} - 2 x^{2} - 5 x - 4 \implies f'=3 x^{2} - 4 x - 5$. This leaves g as $g = \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right)$ which also requires the product rule. Pushing down in the new product rule $f=5 x^{3} - 6 x^{2} + 6 x - 9 \implies f'=15 x^{2} - 12 x + 6$ and $g=- 5 x^{3} - x^{2} - 6 x + 7 \implies g'=- 15 x^{2} - 2 x - 6$. Popping up a level gives $g'=(- 5 x^{3} - x^{2} - 6 x + 7)(15 x^{2} - 12 x + 6)+(5 x^{3} - 6 x^{2} + 6 x - 9)(- 15 x^{2} - 2 x - 6)$Popping up again (Back to the original problem) gives $f'=(x^{3} - 2 x^{2} - 5 x - 4)(\left(- 15 x^{2} - 2 x - 6\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(15 x^{2} - 12 x + 6\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right))+(\left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right))(3 x^{2} - 4 x - 5)=\left(- 15 x^{2} - 2 x - 6\right) \left(x^{3} - 2 x^{2} - 5 x - 4\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(3 x^{2} - 4 x - 5\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(15 x^{2} - 12 x + 6\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(x^{3} - 2 x^{2} - 5 x - 4\right)$}
\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 = (x^{3} - 2 x^{2} - 5 x - 4)(5 x^{3} - 6 x^{2} + 6 x - 9)(- 5 x^{3} - x^{2} - 6 x + 7) " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%28x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20-%204%29%285%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%29%28-%205%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%206%20x%20%2B%207%29%20" alt="LaTeX: \displaystyle y = (x^{3} - 2 x^{2} - 5 x - 4)(5 x^{3} - 6 x^{2} + 6 x - 9)(- 5 x^{3} - x^{2} - 6 x + 7) " data-equation-content=" \displaystyle y = (x^{3} - 2 x^{2} - 5 x - 4)(5 x^{3} - 6 x^{2} + 6 x - 9)(- 5 x^{3} - x^{2} - 6 x + 7) " /> .</p> </p><p> <p>Identifying <img class="equation_image" title=" \displaystyle f=x^{3} - 2 x^{2} - 5 x - 4 " src="/equation_images/%20%5Cdisplaystyle%20f%3Dx%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20-%204%20" alt="LaTeX: \displaystyle f=x^{3} - 2 x^{2} - 5 x - 4 " data-equation-content=" \displaystyle f=x^{3} - 2 x^{2} - 5 x - 4 " /> and <img class="equation_image" title=" \displaystyle g=\left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) " src="/equation_images/%20%5Cdisplaystyle%20g%3D%5Cleft%28-%205%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%206%20x%20%2B%207%5Cright%29%20%5Cleft%285%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%5Cright%29%20" alt="LaTeX: \displaystyle g=\left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) " data-equation-content=" \displaystyle g=\left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) " /> and using the product rule with <img class="equation_image" title=" \displaystyle f=x^{3} - 2 x^{2} - 5 x - 4 \implies f'=3 x^{2} - 4 x - 5 " src="/equation_images/%20%5Cdisplaystyle%20f%3Dx%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20-%204%20%5Cimplies%20f%27%3D3%20x%5E%7B2%7D%20-%204%20x%20-%205%20" alt="LaTeX: \displaystyle f=x^{3} - 2 x^{2} - 5 x - 4 \implies f'=3 x^{2} - 4 x - 5 " data-equation-content=" \displaystyle f=x^{3} - 2 x^{2} - 5 x - 4 \implies f'=3 x^{2} - 4 x - 5 " /> . This leaves g as <img class="equation_image" title=" \displaystyle g = \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cleft%28-%205%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%206%20x%20%2B%207%5Cright%29%20%5Cleft%285%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%5Cright%29%20" alt="LaTeX: \displaystyle g = \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) " data-equation-content=" \displaystyle g = \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) " /> which also requires the product rule. Pushing down in the new product rule <img class="equation_image" title=" \displaystyle f=5 x^{3} - 6 x^{2} + 6 x - 9 \implies f'=15 x^{2} - 12 x + 6 " src="/equation_images/%20%5Cdisplaystyle%20f%3D5%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%20%5Cimplies%20f%27%3D15%20x%5E%7B2%7D%20-%2012%20x%20%2B%206%20" alt="LaTeX: \displaystyle f=5 x^{3} - 6 x^{2} + 6 x - 9 \implies f'=15 x^{2} - 12 x + 6 " data-equation-content=" \displaystyle f=5 x^{3} - 6 x^{2} + 6 x - 9 \implies f'=15 x^{2} - 12 x + 6 " /> and <img class="equation_image" title=" \displaystyle g=- 5 x^{3} - x^{2} - 6 x + 7 \implies g'=- 15 x^{2} - 2 x - 6 " src="/equation_images/%20%5Cdisplaystyle%20g%3D-%205%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%206%20x%20%2B%207%20%5Cimplies%20g%27%3D-%2015%20x%5E%7B2%7D%20-%202%20x%20-%206%20" alt="LaTeX: \displaystyle g=- 5 x^{3} - x^{2} - 6 x + 7 \implies g'=- 15 x^{2} - 2 x - 6 " data-equation-content=" \displaystyle g=- 5 x^{3} - x^{2} - 6 x + 7 \implies g'=- 15 x^{2} - 2 x - 6 " /> . Popping up a level gives <img class="equation_image" title=" \displaystyle g'=(- 5 x^{3} - x^{2} - 6 x + 7)(15 x^{2} - 12 x + 6)+(5 x^{3} - 6 x^{2} + 6 x - 9)(- 15 x^{2} - 2 x - 6) " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%28-%205%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%206%20x%20%2B%207%29%2815%20x%5E%7B2%7D%20-%2012%20x%20%2B%206%29%2B%285%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%29%28-%2015%20x%5E%7B2%7D%20-%202%20x%20-%206%29%20" alt="LaTeX: \displaystyle g'=(- 5 x^{3} - x^{2} - 6 x + 7)(15 x^{2} - 12 x + 6)+(5 x^{3} - 6 x^{2} + 6 x - 9)(- 15 x^{2} - 2 x - 6) " data-equation-content=" \displaystyle g'=(- 5 x^{3} - x^{2} - 6 x + 7)(15 x^{2} - 12 x + 6)+(5 x^{3} - 6 x^{2} + 6 x - 9)(- 15 x^{2} - 2 x - 6) " /> Popping up again (Back to the original problem) gives <img class="equation_image" title=" \displaystyle f'=(x^{3} - 2 x^{2} - 5 x - 4)(\left(- 15 x^{2} - 2 x - 6\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(15 x^{2} - 12 x + 6\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right))+(\left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right))(3 x^{2} - 4 x - 5)=\left(- 15 x^{2} - 2 x - 6\right) \left(x^{3} - 2 x^{2} - 5 x - 4\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(3 x^{2} - 4 x - 5\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(15 x^{2} - 12 x + 6\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(x^{3} - 2 x^{2} - 5 x - 4\right) " src="/equation_images/%20%5Cdisplaystyle%20f%27%3D%28x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20-%204%29%28%5Cleft%28-%2015%20x%5E%7B2%7D%20-%202%20x%20-%206%5Cright%29%20%5Cleft%285%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%5Cright%29%20%2B%20%5Cleft%2815%20x%5E%7B2%7D%20-%2012%20x%20%2B%206%5Cright%29%20%5Cleft%28-%205%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%206%20x%20%2B%207%5Cright%29%29%2B%28%5Cleft%28-%205%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%206%20x%20%2B%207%5Cright%29%20%5Cleft%285%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%5Cright%29%29%283%20x%5E%7B2%7D%20-%204%20x%20-%205%29%3D%5Cleft%28-%2015%20x%5E%7B2%7D%20-%202%20x%20-%206%5Cright%29%20%5Cleft%28x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20-%204%5Cright%29%20%5Cleft%285%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%5Cright%29%20%2B%20%5Cleft%283%20x%5E%7B2%7D%20-%204%20x%20-%205%5Cright%29%20%5Cleft%28-%205%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%206%20x%20%2B%207%5Cright%29%20%5Cleft%285%20x%5E%7B3%7D%20-%206%20x%5E%7B2%7D%20%2B%206%20x%20-%209%5Cright%29%20%2B%20%5Cleft%2815%20x%5E%7B2%7D%20-%2012%20x%20%2B%206%5Cright%29%20%5Cleft%28-%205%20x%5E%7B3%7D%20-%20x%5E%7B2%7D%20-%206%20x%20%2B%207%5Cright%29%20%5Cleft%28x%5E%7B3%7D%20-%202%20x%5E%7B2%7D%20-%205%20x%20-%204%5Cright%29%20" alt="LaTeX: \displaystyle f'=(x^{3} - 2 x^{2} - 5 x - 4)(\left(- 15 x^{2} - 2 x - 6\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(15 x^{2} - 12 x + 6\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right))+(\left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right))(3 x^{2} - 4 x - 5)=\left(- 15 x^{2} - 2 x - 6\right) \left(x^{3} - 2 x^{2} - 5 x - 4\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(3 x^{2} - 4 x - 5\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(15 x^{2} - 12 x + 6\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(x^{3} - 2 x^{2} - 5 x - 4\right) " data-equation-content=" \displaystyle f'=(x^{3} - 2 x^{2} - 5 x - 4)(\left(- 15 x^{2} - 2 x - 6\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(15 x^{2} - 12 x + 6\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right))+(\left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right))(3 x^{2} - 4 x - 5)=\left(- 15 x^{2} - 2 x - 6\right) \left(x^{3} - 2 x^{2} - 5 x - 4\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(3 x^{2} - 4 x - 5\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(5 x^{3} - 6 x^{2} + 6 x - 9\right) + \left(15 x^{2} - 12 x + 6\right) \left(- 5 x^{3} - x^{2} - 6 x + 7\right) \left(x^{3} - 2 x^{2} - 5 x - 4\right) " /> </p> </p>