\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Find the derivative of \(\displaystyle y = \left(- 5 x^{2} + 3 x - 6\right) \ln{\left(x \right)}\).
Using the product rule with \(\displaystyle f = - 5 x^{2} + 3 x - 6\), \(\displaystyle f' = 3 - 10 x\), \(\displaystyle g = \ln{\left(x \right)}\), and \(\displaystyle g'= \frac{1}{x}\) gives:
\begin{equation*} y' = (\ln{\left(x \right)})(3 - 10 x) + (- 5 x^{2} + 3 x - 6)(\frac{1}{x}) = \left(3 - 10 x\right) \ln{\left(x \right)} + \frac{- 5 x^{2} + 3 x - 6}{x} \end{equation*}
\begin{question}Find the derivative of $y = \left(- 5 x^{2} + 3 x - 6\right) \ln{\left(x \right)}$.
\soln{9cm}{Using the product rule with $f = - 5 x^{2} + 3 x - 6$, $f' = 3 - 10 x$, $g = \ln{\left(x \right)}$, and $g'= \frac{1}{x}$ gives:\newline
\begin{equation*} y' = (\ln{\left(x \right)})(3 - 10 x) + (- 5 x^{2} + 3 x - 6)(\frac{1}{x}) = \left(3 - 10 x\right) \ln{\left(x \right)} + \frac{- 5 x^{2} + 3 x - 6}{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 = \left(- 5 x^{2} + 3 x - 6\right) \ln{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%5Cleft%28-%205%20x%5E%7B2%7D%20%2B%203%20x%20-%206%5Cright%29%20%5Cln%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle y = \left(- 5 x^{2} + 3 x - 6\right) \ln{\left(x \right)} " data-equation-content=" \displaystyle y = \left(- 5 x^{2} + 3 x - 6\right) \ln{\left(x \right)} " /> . </p> </p><p> <p>Using the product rule with <img class="equation_image" title=" \displaystyle f = - 5 x^{2} + 3 x - 6 " src="/equation_images/%20%5Cdisplaystyle%20f%20%3D%20-%205%20x%5E%7B2%7D%20%2B%203%20x%20-%206%20" alt="LaTeX: \displaystyle f = - 5 x^{2} + 3 x - 6 " data-equation-content=" \displaystyle f = - 5 x^{2} + 3 x - 6 " /> , <img class="equation_image" title=" \displaystyle f' = 3 - 10 x " src="/equation_images/%20%5Cdisplaystyle%20f%27%20%3D%203%20-%2010%20x%20" alt="LaTeX: \displaystyle f' = 3 - 10 x " data-equation-content=" \displaystyle f' = 3 - 10 x " /> , <img class="equation_image" title=" \displaystyle g = \ln{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20g%20%3D%20%5Cln%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle g = \ln{\left(x \right)} " data-equation-content=" \displaystyle g = \ln{\left(x \right)} " /> , and <img class="equation_image" title=" \displaystyle g'= \frac{1}{x} " src="/equation_images/%20%5Cdisplaystyle%20g%27%3D%20%5Cfrac%7B1%7D%7Bx%7D%20" alt="LaTeX: \displaystyle g'= \frac{1}{x} " data-equation-content=" \displaystyle g'= \frac{1}{x} " /> gives:<br>
<img class="equation_image" title=" y' = (\ln{\left(x \right)})(3 - 10 x) + (- 5 x^{2} + 3 x - 6)(\frac{1}{x}) = \left(3 - 10 x\right) \ln{\left(x \right)} + \frac{- 5 x^{2} + 3 x - 6}{x} " src="/equation_images/%20%20y%27%20%3D%20%20%28%5Cln%7B%5Cleft%28x%20%5Cright%29%7D%29%283%20-%2010%20x%29%20%2B%20%28-%205%20x%5E%7B2%7D%20%2B%203%20x%20-%206%29%28%5Cfrac%7B1%7D%7Bx%7D%29%20%3D%20%5Cleft%283%20-%2010%20x%5Cright%29%20%5Cln%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B-%205%20x%5E%7B2%7D%20%2B%203%20x%20-%206%7D%7Bx%7D%20%20" alt="LaTeX: y' = (\ln{\left(x \right)})(3 - 10 x) + (- 5 x^{2} + 3 x - 6)(\frac{1}{x}) = \left(3 - 10 x\right) \ln{\left(x \right)} + \frac{- 5 x^{2} + 3 x - 6}{x} " data-equation-content=" y' = (\ln{\left(x \right)})(3 - 10 x) + (- 5 x^{2} + 3 x - 6)(\frac{1}{x}) = \left(3 - 10 x\right) \ln{\left(x \right)} + \frac{- 5 x^{2} + 3 x - 6}{x} " /> </p> </p>