\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.5in,right=0.5in,top=0.5in,bottom=0.5in} \pagestyle{fancy} \lhead{\color{blue} \Large Lecture Notes} \chead{\color{black} \LARGE Differentiation 3} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ copyright Hidegkuti 2013} \rfoot{\small Last revised: March 27, 2014} \textwidth 7.6in \textheight 9.6in \setlength{\headheight}{28pt} \setlength{\parindent}{0in} \begin{document} Differentiate each of the following functions.\bigskip \begin{enumerate} \item %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion $f\left( x\right) =\dfrac{\sin x}{x}$ \item $f\left( x\right) =\dfrac{x^{4}-x^{2}+1}{\cos x}$ \item $f\left( x\right) =\dfrac{x+1}{x-1}$ \item $f\left( x\right) =\dfrac{x^{3}-1}{x^{2}}$ \item $f\left( x\right) =\dfrac{1}{x^{2}+1}$ \item $f\left( x\right) =\dfrac{\ln x}{x}$ \item $f\left( \theta \right) =\tan \theta $ \item $f\left( \theta \right) =\sec \theta $ \item $f\left( x\right) =\dfrac{\cos x}{\log _{3}x}$ \item $f\left( x\right) =\dfrac{\sqrt{x}}{x^{3}}$ \item $f\left( x\right) =\log _{2}x+\log _{x}2$ \item $f\left( x\right) =\dfrac{1+\ln x}{x^{2}-\ln x}$% %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \item Preview of calculus 2. \ A bit more on tangent and secant. a) \ Prove the identity $\tan ^{2}x+1=\sec ^{2}x$ b) \ Based on the identity above, re-write $f^{\prime }\left( x\right) $ when $f\left( x\right) =\tan x$. c) \ Based on the previous answer, find $\dint \tan ^{2}xdx$ d) \ Compute $f^{\prime }\left( x\right) $ if $f\left( x\right) =\sec x$ and re-write it in terms of tangent and/or secant. \pagebreak \end{enumerate} \begin{center} {\LARGE Answers\bigskip } \end{center} \bigskip %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion 1.) $\ f^{\prime }\left( x\right) =\dfrac{x\cos x-\sin x}{x^{2}}\qquad $2.) $% \ f^{\prime }\left( x\right) =\dfrac{\left( 4x^{3}-2x\right) \cos x+\left( x^{4}-x^{2}+1\right) \sin x}{\cos ^{2}x}\qquad $3.) $\ f^{\prime }\left( x\right) =-\dfrac{2}{\left( x-1\right) ^{2}}$\bigskip 4.) $\ f^{\prime }\left( x\right) =1+\dfrac{2}{x^{3}}$ or \ $\dfrac{x^{3}+2}{% x^{3}}\qquad $5.) $\ f^{\prime }\left( x\right) =\dfrac{-2x}{\left( x^{2}+1\right) ^{2}}\qquad $6.) $\ f^{\prime }\left( x\right) =\dfrac{1-\ln x% }{x^{2}}\qquad $7.) $\ f^{\prime }\left( \theta \right) =\dfrac{1}{\cos ^{2}\theta }$\bigskip 8.) $\ f^{\prime }\left( \theta \right) =\dfrac{\sin \theta }{\cos ^{2}\theta }\bigskip $ Note that the result can also be written as $\dfrac{\sin \theta }{\cos ^{2}\theta }=\dfrac{\sin \theta }{\cos \theta }\cdot \dfrac{1}{\cos \theta }% =\tan \theta \sec \theta \bigskip $ 9.) $\ f^{\prime }\left( x\right) =\dfrac{-\sin x\log _{3}x-\cos x\dfrac{1}{% x\ln 3}}{\left( \log _{3}x\right) ^{2}}=\dfrac{-x\ln 3\sin x\log _{3}x-\cos x% }{\left( \log _{3}x\right) ^{2}x\ln 3}=\dfrac{-x\sin x\ln x-\cos x}{\left( \log _{3}x\right) ^{2}x\ln 3}\bigskip $ $\ \ \ \ \ \ \ \ \ \ =\dfrac{-x\sin x\ln x-\cos x}{\left( \dfrac{\ln x}{\ln 3% }\right) ^{2}x\ln 3}=\dfrac{-\ln 3\left( x\ln x\sin x+\cos x\right) }{% x\left( \ln x\right) ^{2}}\bigskip $ 10.) $\ f^{\prime }\left( x\right) =-\dfrac{5\sqrt{x}}{2x^{4}}$\bigskip\ \qquad \qquad 11.) $\ f^{\prime }\left( x\right) =\dfrac{1}{x\ln 2}-\dfrac{% \ln 2}{x\ln ^{2}x}\qquad $12.) $\ f^{\prime }\left( x\right) =\dfrac{% -x^{2}-2x^{2}\ln x+1}{x\left( x^{2}-\ln x\right) ^{2}}$\bigskip 13.) \ a) \ $\tan ^{2}x+1=\sec ^{2}x$% \begin{equation*} \text{LHS}=\tan ^{2}x+1=\dfrac{\sin ^{2}x}{\cos ^{2}x}+1=\dfrac{\sin ^{2}x}{% \cos ^{2}x}+\dfrac{\cos ^{2}x}{\cos ^{2}x}=\dfrac{\sin ^{2}x+\cos ^{2}x}{% \cos ^{2}x}=\dfrac{1}{\cos ^{2}x}=\sec ^{2}x=\text{RHS} \end{equation*} \qquad b) \ $f^{\prime }\left( x\right) =\dfrac{1}{\cos ^{2}x}=\sec ^{2}x=1+\tan ^{2}x$ \ \qquad c) \ $\dint \tan ^{2}xdx=\tan x-x+C$\bigskip\ \qquad d) \ $f^{\prime }\left( x\right) =\sec x\tan x$ \bigskip \bigskip \vspace{2in} \bigskip \bigskip \bigskip \bigskip {\small %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}