%% This document created by Scientific Notebook (R) Version 3.5 %% Starting shell: article \documentclass{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amssymb} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2570} %TCIDATA{} %TCIDATA{Created=Wednesday, February 10, 1999 13:29:48} %TCIDATA{LastRevised=Sunday, February 13, 2005 15:49:44} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=On line bluem.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{Counters=arabic,1} %TCIDATA{AllPages= %H=36 %F=36,\PARA{038
\QTR{small}{Matematick\U{e1} anal\U{fd}za I online - Diferenci\U{e1}lny po\U{10d}et funkci\U{ed} jednej re\U{e1}lnej premennej - Diferenci\U{e1}l funkcie a pravidl\U{e1} derivovania\dotfill \thepage }} %} \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}[theorem]{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]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \author{A. U. Thor} \title{Lab Report} \date{The Date } \maketitle \begin{abstract} A Laboratory report created with Scientific Notebook \end{abstract} \section{Diferenci\'{a}lny po\v{c}et funkci\'{\i} jednej re\'{a}lnej premennej} \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{maindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{maindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{M5.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{M5.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{M52.tex}}% %BeginExpansion \msihyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{M52.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{M54.tex}}% %BeginExpansion \msihyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{M54.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{O5.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{O5.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Cvi\v{c}enia}{}{}{C5.tex}}% %BeginExpansion \msihyperref{Cvi\v{c}enia}{}{}{C5.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{G1.tex}}% %BeginExpansion \msihyperref{Index}{}{}{G1.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \subsection{Diferenci\'{a}l funkcie a pravidl\'{a} derivovania funkci\'{\i}} Ak je $f:A\longrightarrow \mathbf{R}$ diferencovate\v{l}n\'{a} v bode $a\in A $, t.j. plat\'{\i} vz\v{t}ah $f(x)-f(a)=f\,$\/$\,^{\prime }(a)(x-a)+r(x)(x-a) $, v ktorom pou\v{z}ijeme substit\'{u}ciu $x=a+h$, dostaneme $f(a+h)-f(a)=f\, $\/$\,^{\prime }(a)h+r(a+h)h.$ Preto\v{z}e plat% \'{\i} $\lim_{h\longrightarrow 0}r\left( a+h\right) =0$, tak m\'{a}me \[ f(a+h)-f(a)\approx f\,\/\,^{\prime }(a)h. \]% V\'{y}raz $f\,$\/$\,^{\prime }(a)h$ sa naz\'{y}va \label{3}\emph{diferenci% \'{a}l} funkcie $f$ v bode $a$ a ozna\v{c}uje sa $df_{a}$. M\'{a}me teda $% df_{a}=f\,$\/$\,^{\prime }(a)h.$ Ak $g(x)=x$, tak \[ dg_{a}=dx_{a}=g^{\prime }(a)h=h,\,\forall a\in \mathbf{R}. \]% Teda \[ df_{a}=f\,\/\,^{\prime }(a)dx_{a} \]% zvy\v{c}ajne v\v{s}ak nep\'{\i}\v{s}eme diferenci\'{a}l v bode $a$, ale v bode $x$, t.j. \[ df_{x}=df=f\,\/\,^{\prime }(x)dx. \]% Preto\v{z}e $f(a+h)\approx f(a)+df_{a}$ vieme aproximova\v{t} hodnotu funkcie $f\left( a+h\right) $ pomocou hodnoty funkcie v bode $a$ t.j. $% f\left( a\right) $ a diferenci\'{a}lu $df_{a}$. Diferenci\'{a}l m\'{a} elegantn\'{e} vyjadrenie v Leibnitzovom ozna\v{c}en% \'{\i}: \[ df=\frac{df}{dx}dx. \] \begin{theorem} \label{4}(Veta o pravidl\'{a}ch derivovania)\ Nech $f,g:A\longrightarrow \mathbf{R}$ s\'{u} diferencovate\v{l}n\'{e} v bode $a\in A,$ nech $\,c\in \mathbf{R.}$ Potom% \[ (cf)^{\prime }\left( a\right) =cf\,\/\,^{\prime }\left( a\right) , \]% \[ (f+g)^{\prime }\left( a\right) =f\,\/\,^{\prime }\left( a\right) +g^{\prime }\left( a\right) , \]% \[ (fg)^{\prime }\left( a\right) =f\,\/\,^{\prime }\left( a\right) g\left( a\right) +f\left( a\right) g^{\prime }\left( a\right) . \]% Ak $\,g\left( a\right) \neq 0,$ potom \[ \left( \frac{f}{g}\right) ^{\prime }\left( a\right) =\frac{f\,\/\,^{\prime }\left( a\right) g\left( a\right) -f\left( a\right) g^{\prime }\left( a\right) }{g^{2}\left( a\right) }. \] \end{theorem} $% \begin{tabular}{|c|} \hline {\small %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DO531.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DO531.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{1}$ \begin{example} Nech $f:\mathbf{R}\longrightarrow \mathbf{R},\,f\left( x\right) =x\sin x$. N% \'{a}jdite $f\,$\/$\,^{\prime }(x)$ pre \v{l}ubovo\v{l}n\'{e} $x\in \mathbf{R% }$. \end{example} \begin{solution} Pou\v{z}ijeme vetu o deriv\'{a}cii s\'{u}\v{c}inu dvoch funkci\'{\i}: \[ \left( x\sin x\right) ^{\prime }=\left( x\right) ^{\prime }\sin x+x\left( \sin x\right) ^{\prime }=\sin x+x\cos x.\,\square \] \end{solution} \begin{example} Nech $f:\mathbf{R}\longrightarrow \mathbf{R},\,f\left( x\right) =x^{n},\,n\in \mathbf{N}$. Dok\'{a}\v{z}te, \v{z}e $f\,$\/$\,^{\prime }(x)=nx^{n-1}$ pre ka\v{z}d\'{e} $x\in \mathbf{R}$. \end{example} \begin{solution} \[ f\,\/\,^{\prime }(x)=\left( x^{n}\right) ^{\prime }=\lim_{t\longrightarrow x}% \frac{t^{n}-x^{n}}{t-x}=\lim_{t\longrightarrow x}\frac{\left( t-x\right) \left( t^{n-1}+t^{n-2}x+...+tx^{n-2}+x^{n-1}\right) }{t-x}% =nx^{n-1}.\,\square \bigskip \] \end{solution} \begin{example} Uk\'{a}\v{z}te, \v{z}e $(x^{-n})^{\prime }=-nx^{-n-1},\,n\in \mathbf{N},$\ pre ka\v{z}d\'{e} $x\in \mathbf{R\setminus }\left\{ 0\right\} .$ \end{example} \begin{solution} Pou\v{z}it\'{\i}m vety o deriv\'{a}cii podielu dvoch funkci\'{\i} m\'{a}me:% \[ (x^{-n})^{\prime }=\left( \frac{1}{x^{n}}\right) ^{\prime }=\frac{\left( 1\right) ^{\prime }x^{n}-1\left( x^{n}\right) ^{\prime }}{\left( x^{n}\right) ^{2}}=\frac{-nx^{n-1}}{x^{2n}}=-nx^{-n-1}. \]% Pre $n=0,\,\,f\left( x\right) =x^{0}=1$ plat\'{\i} $f\,$\/$\,^{\prime }\left( x\right) =0.\,\square $ \end{solution} \begin{description} \item[Pozn\'{a}mka] Na z\'{a}klade v\'{y}sledkov predch\'{a}dzaj\'{u}cich dvoch pr\'{\i}kladov m\^{o}\v{z}eme p\'{\i}sa\v{t}: \[ \left( x^{n}\right) ^{\prime }=nx^{n-1},\,\forall n\in \mathbf{Z},\ \forall x\in \mathbf{R\setminus }\left\{ 0\right\} . \] \end{description} \begin{example} Uk\'{a}\v{z}te, \v{z}e plat\'{\i} \[ (c_{n}x^{n}+c_{n-1}x^{n-1}+...+c_{1}x+c_{0})^{\prime }=nc_{n}x^{n-1}+(n-1)c_{n-1}x^{n-2}+...+2c_{2}x+c_{1}. \] \end{example} \begin{solution} Pou\v{z}it\'{\i}m vety o deriv\'{a}cii n\'{a}sobku funkcie re\'{a}lnym \v{c}% \'{\i}slom a vety o deriv\'{a}cii s\'{u}\v{c}tu dvoch funkci\'{\i} dostaneme:% $\bigskip $% \[ (c_{n}x^{n}+c_{n-1}x^{n-1}+...+c_{1}x+c_{0})^{\prime }=c_{n}(x^{n})^{\prime }+c_{n-1}\left( x^{n-1}\right) ^{\prime }+...+c_{1}\left( x\right) ^{\prime }+c_{0}\left( 1\right) ^{\prime }= \]% \[ =nc_{n}x^{n-1}+(n-1)c_{n-1}x^{n-2}+...+2c_{2}x+c_{1}.\,\square \] \end{solution} \begin{example} Nech $h:\mathbf{R}\longrightarrow \mathbf{R},\,h\left( x\right) =\frac{5x^{6}% }{x^{4}+1}$. N\'{a}jdite $h^{\prime }\left( x\right) $. \end{example} \begin{solution} Pou\v{z}it\'{\i}m vety o deriv\'{a}cii podielu dvoch funkci\'{\i} m\'{a}me: \[ h^{\prime }\left( x\right) =\left( \frac{5x^{6}}{x^{4}+1}\right) ^{\prime }=% \frac{\left( 5x^{6}\right) ^{\prime }\left( x^{4}+1\right) -\left( 5x^{6}\right) \left( x^{4}+1\right) ^{\prime }}{\left( x^{4}+1\right) ^{2}}= \]% \[ =\frac{30x^{5}\left( x^{4}+1\right) -\left( 5x^{6}\right) \left( 4x^{3}\right) }{\left( x^{4}+1\right) ^{2}}=\frac{10x^{9}+30x^{5}}{\left( x^{4}+1\right) ^{2}}.\,\square \] \end{solution} \begin{theorem} \label{5}(Veta o deriv\'{a}cii zlo\v{z}enej funkcie) Nech \[ f:B\longrightarrow \mathbf{R},\,B\subset \mathbf{R},\,g:A\longrightarrow \mathbf{R},\,A\subset \mathbf{R} \]% tak, \v{z}e $\emptyset \subset H\left( g\right) \subset D\left( f\right) $ a zlo\v{z}en\'{a} funkcia \[ F=f\circ g:\left\{ x\in A:g\left( x\right) \in B\right\} \longrightarrow \mathbf{R} \]% je definovan\'{a} v okol\'{\i} bodu $a$. Nech funkcia $g$ je diferencovate% \v{l}n\'{a} v bode $a\in A$ a m\'{a} deriv\'{a}ciu $g^{\prime }\left( a\right) $ a nech funkcia $f$ je diferencovate\v{l}n\'{a} v bode $b=g\left( a\right) $ a m\'{a} deriv\'{a}ciu $f\,$\/$\,^{\prime }\left( b\right) $. Potom zlo\v{z}en\'{a} funkcia $F=f\circ g$ je diferencovate\v{l}n\'{a} v bode $a$ a m\'{a} deriv\'{a}ciu \[ F^{\prime }\left( a\right) =f\,\/\,^{\prime }\left( g\left( a\right) \right) g^{\prime }\left( a\right) . \] \end{theorem} $% \begin{tabular}{|c|} \hline %TCIMACRO{\hyperref{{\small D\^{o}kaz}}{}{}{DO532.tex} }% %BeginExpansion \msihyperref{{\small D\^{o}kaz}}{}{}{DO532.tex} %EndExpansion \\ \hline \end{tabular}% \label{2}$ \begin{description} \item[Pozn\'{a}mka] Ak s\'{u} splnen\'{e} predpoklady vety o pravidl\'{a}ch derivovania a vety o deriv\'{a}cii zlo\v{z}enej funkcie zvykneme ich zapisova% \v{t} skr\'{a}tene bez vypisovania argumentov takto:% \[ (cf)^{\prime }=cf\,\/\,^{\prime }, \]% \[ (f+g)^{\prime }=f\,\/\,^{\prime }+g^{\prime }, \]% \[ (fg)^{\prime }=f\,\/\,^{\prime }g+fg^{\prime }, \]% \[ \left( \frac{f}{g}\right) ^{\prime }=\frac{f\,\/\,^{\prime }g-fg^{\prime }}{% g^{2}}, \]% \[ \left( f\circ g\right) ^{\prime }=\left( f\,\/\,^{\prime }\circ g\right) g^{\prime }. \] \end{description} \begin{example} Nech $h:\mathbf{R}\longrightarrow \mathbf{R,}h\left( x\right) =\sin 3x.$N% \'{a}jdite $h^{\prime }\left( x\right) $. \end{example} \begin{solution} Nech $u=g(x)=3x,\,f\left( u\right) =\sin u\Longrightarrow h=f\circ g$. Potom pod\v{l}a vety o deriv\'{a}cii zlo\v{z}enej funkcie plat\'{\i} $h^{\prime }\left( x\right) =\left[ f\,\/\,^{\prime }\left( u\right) \right] _{u=3x}g^{\prime }\left( x\right) =\left[ \cos u\right] _{u=3x}3=3\cos 3x.\square $ \end{solution} \begin{example} Nech $f:\mathbf{R}\longrightarrow \mathbf{R},\,f\left( x\right) =\sqrt{% 1+x^{4}}$. N\'{a}jdite $f\,$\/$\,^{\prime }$. \end{example} \begin{solution} Pod\v{l}a vety o deriv\'{a}cii zlo\v{z}enej funkcie m\'{a}me:% \[ \left( \sqrt{1+x^{4}}\right) ^{\prime }=\left( \left( 1+x^{4}\right) ^{\frac{% 1}{2}}\right) ^{\prime }=\frac{1}{2}\left( 1+x^{4}\right) ^{-\frac{1}{2}% }\left( 1+x^{4}\right) ^{\prime }= \]% \[ =\frac{1}{2}\left( 1+x^{4}\right) ^{-\frac{1}{2}}4x^{3}=\frac{2x^{3}}{\sqrt{% 1+x^{4}}}.\square \] \end{solution} Derivovanie zlo\v{z}enej funkcie je ve\v{l}mi jednoduch\'{e} pri pou\v{z}it% \'{\i} Leibnitzovho ozna\v{c}enia. Podobne ako vo vete o deriv\'{a}cii zlo% \v{z}enej funkcie polo\v{z}me $y=(f\circ g)(x)$ a ozna\v{c}me $% u=g(x),\,y=f(u)$, potom $y=f(g(x))$, $\frac{du}{dx}=g^{\prime }\left( x\right) $, $\frac{dy}{du}=f\,$\/$\,^{\prime }\left( u\right) $. Tak dostaneme \[ \frac{dy}{dx}=\frac{d}{dx}f\left( g\left( x\right) \right) =f\,\/\,^{\prime }\left( g\left( x\right) \right) g^{\prime }\left( x\right) =f\,\/\,^{\prime }\left( u\right) g^{\prime }\left( x\right) =\frac{dy}{du}\frac{du}{dx}. \] \begin{example} Uk\'{a}\v{z}te, \v{z}e $\left( x^{p}\right) ^{\prime }=px^{p-1}$ pre ka\v{z}d% \'{e} $p\in \mathbf{Q}$. \end{example} \begin{solution} Tvrdenie sme u\v{z} uk\'{a}zali pre $n\in \mathbf{N}$. Najsk\^{o}r priamo z defin\'{\i}cie uk\'{a}\v{z}eme, \v{z}e ak $m\in \mathbf{N}$, tak pre funkciu $f(x)=x^{\frac{1}{m}}$ plat\'{\i} $f\,$\/$\,^{\prime }\left( x\right) =\frac{% 1}{m}x^{\frac{1}{m}-1}$: \[ \left( x^{\frac{1}{m}}\right) ^{\prime }=\lim_{t\longrightarrow x}\frac{t^{% \frac{1}{m}}-x^{\frac{1}{m}}}{t-x}=\lim_{t\longrightarrow x}\frac{t^{\frac{1% }{m}}-x^{\frac{1}{m}}}{\left( t^{\frac{1}{m}}\right) ^{m}-\left( x^{\frac{1}{% m}}\right) ^{m}}= \]% \[ =\lim_{t\longrightarrow x}\frac{t^{\frac{1}{m}}-x^{\frac{1}{m}}}{\left( t^{% \frac{1}{m}}-x^{\frac{1}{m}}\right) \left( \left( t^{\frac{1}{m}}\right) ^{m-1}+\left( t^{\frac{1}{m}}\right) ^{m-2}x^{\frac{1}{m}}+...+t^{\frac{1}{m}% }\left( x^{\frac{1}{m}}\right) ^{m-2}+\left( x^{\frac{1}{m}}\right) ^{m-1}\right) }= \]% \[ =\frac{1}{\left( \left( x^{\frac{1}{m}}\right) ^{m-1}+\left( x^{\frac{1}{m}% }\right) ^{m-2}x^{\frac{1}{m}}+...+x^{\frac{1}{m}}\left( x^{\frac{1}{m}% }\right) ^{m-2}+\left( x^{\frac{1}{m}}\right) ^{m-1}\right) }=\frac{1}{% m\left( x^{\frac{1}{m}}\right) ^{m-1}}. \]% Potom pre \v{l}ubovo\v{l}n\'{e} racion\'{a}lne \v{c}\'{\i}slo $p$ v tvare $p=% \frac{n}{m},\,n\in \mathbf{Z}$ plat\'{\i}: \[ x^{\frac{n}{m}}=\left( x^{\frac{1}{m}}\right) ^{n} \]% a aplik\'{a}ciou vety o deriv\'{a}cii zlo\v{z}enej funkcie dost\'{a}vame \[ \left( x^{\frac{n}{m}}\right) ^{\prime }=\left( \left( x^{\frac{1}{m}% }\right) ^{n}\right) ^{\prime }=n\left( x^{\frac{1}{m}}\right) ^{n-1}\left( x^{\frac{1}{m}}\right) ^{\prime }=nx^{\frac{n}{m}-\frac{1}{m}}\frac{1}{m}x^{% \frac{1}{m}-1}=\frac{n}{m}x^{\frac{n}{m}-1}.\square \] \end{solution} Vymenujme niektor\'{e} pravidl\'{a} derivovania:% \[ \left( x^{a}\right) ^{\prime }=ax^{a-1},\,a\in \mathbf{R,} \]% \[ \left( \ln x\right) ^{\prime }=\frac{1}{x}, \]% \[ \left( e^{x}\right) ^{\prime }=e^{x}, \]% \[ \left( a^{x}\right) ^{\prime }=a^{x}\ln a,\,a>0, \]% \[ \left( \sin x\right) ^{\prime }=\cos x, \]% \[ \left( \cos x\right) ^{\prime }=-\sin x, \]% \[ \left( \limfunc{tg}x\right) ^{\prime }=\frac{1}{\cos ^{2}x}, \]% \[ \left( \func{cotg}x\right) ^{\prime }=-\frac{1}{\sin ^{2}x}, \]% \[ \left( \limfunc{arctg}x\right) ^{\prime }=\frac{1}{1+x^{2}}, \]% \[ \left( \limfunc{arccotg}x\right) ^{\prime }=-\frac{1}{1+x^{2}}, \]% \[ \left( \arcsin x\right) ^{\prime }=\frac{1}{\sqrt{1-x^{2}}}, \]% \[ \left( \arccos x\right) ^{\prime }=-\frac{1}{\sqrt{1-x^{2}}}, \]% \[ \left( \ln \left| \frac{1+x}{1-x}\right| \right) ^{\prime }=\frac{2}{1-x^{2}}% , \]% \[ \left( \ln \left| x+\sqrt{x^{2}+a^{2}}\right| \right) ^{\prime }=\frac{1}{% \sqrt{x^{2}+a^{2}}}, \]% \[ \left( \ln \left| f\left( x\right) \right| \right) ^{\prime }=\frac{% f\,\/\,^{\prime }\left( x\right) }{f\left( x\right) }. \] \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{maindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{maindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{M5.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{M5.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{M52.tex}}% %BeginExpansion \msihyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{M52.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{M54.tex}}% %BeginExpansion \msihyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{M54.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{O5.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{O5.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Cvi\v{c}enia}{}{}{C5.tex}}% %BeginExpansion \msihyperref{Cvi\v{c}enia}{}{}{C5.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{G1.tex}}% %BeginExpansion \msihyperref{Index}{}{}{G1.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \rule{6.5in}{0.04in} \textsl{Matematick\'{a} anal\'{y}za I} \section{Diferenci\'{a}lny po\v{c}et funkci\'{\i} jednej re\'{a}lnej premennej} \end{document}