\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Find the derivative of \(\displaystyle F(x) = \int\limits_{6}^{\frac{1}{x^{2}}} \frac{1}{t^{2}}\, dt\).
Using the Fundamental Theorem of Calculus part I and the chain rule with \(\displaystyle u=\frac{1}{x^{2}}\) and \(\displaystyle F(u)=\int\limits_{6}^{u} \frac{1}{t^{2}}\, dt\) gives: \(\displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(\frac{1}{u^{2}}\right)\left(- \frac{2}{x^{3}}\right)=- 2 x\)
\begin{question}Find the derivative of $F(x) = \int\limits_{6}^{\frac{1}{x^{2}}} \frac{1}{t^{2}}\, dt$.
\soln{9cm}{Using the Fundamental Theorem of Calculus part I and the chain rule with $u=\frac{1}{x^{2}}$ and $F(u)=\int\limits_{6}^{u} \frac{1}{t^{2}}\, dt$ gives: $F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(\frac{1}{u^{2}}\right)\left(- \frac{2}{x^{3}}\right)=- 2 x$}
\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 F(x) = \int\limits_{6}^{\frac{1}{x^{2}}} \frac{1}{t^{2}}\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28x%29%20%3D%20%5Cint%5Climits_%7B6%7D%5E%7B%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%7D%7D%20%5Cfrac%7B1%7D%7Bt%5E%7B2%7D%7D%5C%2C%20dt%20" alt="LaTeX: \displaystyle F(x) = \int\limits_{6}^{\frac{1}{x^{2}}} \frac{1}{t^{2}}\, dt " data-equation-content=" \displaystyle F(x) = \int\limits_{6}^{\frac{1}{x^{2}}} \frac{1}{t^{2}}\, dt " /> . </p> </p><p> <p>Using the Fundamental Theorem of Calculus part I and the chain rule with <img class="equation_image" title=" \displaystyle u=\frac{1}{x^{2}} " src="/equation_images/%20%5Cdisplaystyle%20u%3D%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%7D%20" alt="LaTeX: \displaystyle u=\frac{1}{x^{2}} " data-equation-content=" \displaystyle u=\frac{1}{x^{2}} " /> and <img class="equation_image" title=" \displaystyle F(u)=\int\limits_{6}^{u} \frac{1}{t^{2}}\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28u%29%3D%5Cint%5Climits_%7B6%7D%5E%7Bu%7D%20%5Cfrac%7B1%7D%7Bt%5E%7B2%7D%7D%5C%2C%20dt%20" alt="LaTeX: \displaystyle F(u)=\int\limits_{6}^{u} \frac{1}{t^{2}}\, dt " data-equation-content=" \displaystyle F(u)=\int\limits_{6}^{u} \frac{1}{t^{2}}\, dt " /> gives: <img class="equation_image" title=" \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(\frac{1}{u^{2}}\right)\left(- \frac{2}{x^{3}}\right)=- 2 x " src="/equation_images/%20%5Cdisplaystyle%20F%27%28x%29%3D%5Cfrac%7BdF%7D%7Bdu%7D%5Cfrac%7Bdu%7D%7Bdx%7D%20%3D%20%5Cleft%28%5Cfrac%7B1%7D%7Bu%5E%7B2%7D%7D%5Cright%29%5Cleft%28-%20%5Cfrac%7B2%7D%7Bx%5E%7B3%7D%7D%5Cright%29%3D-%202%20x%20" alt="LaTeX: \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(\frac{1}{u^{2}}\right)\left(- \frac{2}{x^{3}}\right)=- 2 x " data-equation-content=" \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(\frac{1}{u^{2}}\right)\left(- \frac{2}{x^{3}}\right)=- 2 x " /> </p> </p>