%% 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 16:54:33} %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 II online - Diferencovate\U{13e}n\U{e9} funkcie - Diferencovanie funkci\U{ed}\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{Diferencovate\v{l}n\'{e} funkcie} \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{maiindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{maiindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{Ma3.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{Ma3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{Ma32.tex}}% %BeginExpansion \msihyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{Ma32.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{Ma34.tex}}% %BeginExpansion \msihyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{Ma34.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{O3.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{O3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Cvi\v{c}enia}{}{}{C3.tex}}% %BeginExpansion \msihyperref{Cvi\v{c}enia}{}{}{C3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{Glos.tex}}% %BeginExpansion \msihyperref{Index}{}{}{Glos.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \subsection{\protect\bigskip Diferencovanie funkci\'{\i}.} \begin{theorem} \label{4}(Veta o diferencovan\'{\i} s\'{u}\v{c}tu a n\'{a}sobku funkcie) Nech $\mathbf{f,g}:A\left( \subset \mathbf{R}^{n}\right) \longrightarrow \mathbf{R}^{m}$ s\'{u} diferencovate\v{l}n\'{e} v bode $\mathbf{a}\in G.$ Potom aj funkcie $\mathbf{f}+\mathbf{g},$ $\alpha \mathbf{f}$ $\mathbf{\,}$ (% $\alpha \in \mathbf{R}$) s\'{u} diferencovate\v{l}n\'{e} v bode $\mathbf{a}$ a plat\'{\i} \[ D\left( \mathbf{f}+\mathbf{g}\right) _{\mathbf{a}}=D\mathbf{f}_{\mathbf{a}}+D% \mathbf{g}_{\mathbf{a}}, \]% \[ D\left( \alpha \mathbf{f}\right) _{\mathbf{a}}=\alpha D\mathbf{f}_{\mathbf{a}% }. \] \end{theorem} \begin{tabular}{|c|} \hline {\small %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DO331.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DO331.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{1} Teraz budeme sk\'{u}ma\v{t} zlo\v{z}en\'{e} funkcie. Nech $G\subset \mathbf{R% }^{n}$ a $E\subset \mathbf{R}^{p}$ s\'{u} otvoren\'{e} mno\v{z}iny. Predpokladajme, \v{z}e $\mathbf{f}:E\longrightarrow \mathbf{R}^{m},\,\mathbf{% g}:G\longrightarrow \mathbf{R}^{p}$ s\'{u} tak\'{e}, \v{z}e $\mathbf{g}% (G)\subset E.$ Potom $\mathbf{h}:G\longrightarrow \mathbf{R}^{m},\,\mathbf{h}% (\mathbf{\,x\,})=\left( \mathbf{f\circ g}\right) \left( \mathbf{x}\right) =% \mathbf{f}(\mathbf{g}(\mathbf{\,x\,}))$ je zlo\v{z}en\'{a} funkcia z funkci% \'{\i} $\mathbf{f}$ a $\mathbf{g,}$ \v{c}o zapisujeme $\mathbf{h}=\mathbf{% f\circ g}.$ \begin{theorem} \label{5}(Veta o diferencovan\'{\i} zlo\v{z}enej funkcie) Nech $G\subset \mathbf{R}^{n}$ a $E\subset \mathbf{R}^{p}$ s\'{u} otvoren\'{e} mno\v{z}iny. Predpokladajme, \v{z}e $\mathbf{f}:E\longrightarrow \mathbf{R}^{m},\,\mathbf{% g}:G\longrightarrow \mathbf{R}^{p}$ s\'{u} tak\'{e}, \v{z}e $\mathbf{g}% (G)\subset E.$ Nech funkcia $\mathbf{h}:G\longrightarrow \mathbf{R}^{m},\,% \mathbf{h}(\mathbf{\,x\,})=\left( \mathbf{f\circ g}\right) \left( \mathbf{x}% \right) =\mathbf{f}(\mathbf{g}(\mathbf{\,x\,}))$ je zlo\v{z}en\'{a} funkcia. Ak $\mathbf{g}$ je diferencovate\v{l}n\'{a} v bode $\mathbf{\,x\,}\in G$ a $% \mathbf{f}$ je diferencovate\v{l}n\'{a} v bode $\mathbf{\,y\,}=\mathbf{g}(% \mathbf{\,x\,})\in E,$ potom zlo\v{z}en\'{a} funkcia $\mathbf{h}$ = $\mathbf{% f}$ $\mathbf{\circ }$ $\mathbf{g}$ je diferencovate\v{l}n\'{a} v bode $% \mathbf{\,x\,}\in G$ a plat\'{\i} $D\mathbf{h}_{\mathbf{x}}=D\mathbf{f}_{% \mathbf{y}}D\mathbf{g}_{\mathbf{x}}.$ \end{theorem} \begin{tabular}{|c|} \hline {\small %TCIMACRO{\hyperref{D\^{o}kaz}{}{}{DO332.tex}}% %BeginExpansion \msihyperref{D\^{o}kaz}{}{}{DO332.tex}% %EndExpansion } \\ \hline \end{tabular}% \label{2} \begin{description} \item[Pozn\'{a}mka] $D\mathbf{f}_{\mathbf{y}}D\mathbf{g}_{\mathbf{x}}$ m\^{o}% $\mathbf{\,}$\v{z}me ch\'{a}pa\v{t} bu\v{d} ako zlo\v{z}en\'{y} line\'{a}rny oper\'{a}tor, alebo ako s\'{u}\v{c}in mat\'{\i}c. \end{description} \begin{example} \label{3}Nech $\mathbf{g}:\mathbf{R}^{3}\longrightarrow \mathbf{R}^{3},\,f:% \mathbf{R}^{3}\longrightarrow \mathbf{R}$ s\'{u} diferencovate\v{l}n\'{e} funkcie a nech $h=f\mathbf{\circ g}:\mathbf{R}^{3}\longrightarrow \mathbf{R}% . $ Potom $Dh_{\mathbf{x}}=Df_{\mathbf{g}\left( \mathbf{x}\right) }D\mathbf{g% }_{\mathbf{x}}$ je matica typu $1\times 3,$ ktor\'{u} dostaneme n\'{a}soben% \'{\i}m matice $1\times 3$ - $Df_{\mathbf{g}\left( \mathbf{x}\right) }$ a matice $1\times 3$ - $D\mathbf{g}_{\mathbf{x}}.$ Vyjadren\'{\i}m v komponent% \'{a}ch, potom m\'{a}me \[ \left( \frac{\partial h\left( \mathbf{x}\right) }{\partial x_{1}},\frac{% \partial h\left( \mathbf{x}\right) }{\partial x_{2}},\frac{\partial h\left( \mathbf{x}\right) }{\partial x_{3}}\right) =\left( \frac{\partial f\left( \mathbf{y}\right) }{\partial y_{1}},\frac{\partial f\left( \mathbf{y}\right) }{\partial y_{2}},\frac{\partial f\left( \mathbf{y}\right) }{\partial y_{3}}% \right) \left( \begin{array}{ccc} \frac{\partial g_{1}\left( \mathbf{x}\right) }{\partial x_{1}} & \frac{% \partial g_{1}\left( \mathbf{x}\right) }{\partial x_{2}} & \frac{\partial g_{1}\left( \mathbf{x}\right) }{\partial x_{3}} \\ \frac{\partial g_{2}\left( \mathbf{x}\right) }{\partial x_{1}} & \frac{% \partial g_{2}\left( \mathbf{x}\right) }{\partial x_{2}} & \frac{\partial g_{2}\left( \mathbf{x}\right) }{\partial x_{3}} \\ \frac{\partial g_{3}\left( \mathbf{x}\right) }{\partial x_{1}} & \frac{% \partial g_{3}\left( \mathbf{x}\right) }{\partial x_{2}} & \frac{\partial g_{3}\left( \mathbf{x}\right) }{\partial x_{3}}% \end{array}% \right) \]% $\mathbf{\,}$ teda \[ \frac{\partial h\left( \mathbf{x}\right) }{\partial x_{i}}=\sum_{j=1}^{3}% \frac{\partial f\left( \mathbf{y}\right) }{\partial y_{j}}\frac{\partial g_{j}\left( \mathbf{x}\right) }{\partial x_{i}}, \]% $\mathbf{\,}$ kde $\mathbf{\,y\,}=\mathbf{g}(\mathbf{\,x\,}),\mathbf{\,}% i=1,2,3.$ \v{C}astej\v{s}ie sa pou\v{z}\'{\i}va alternat\'{\i}vne ozna\v{c}% enie \begin{equation} \frac{\partial h}{\partial x_{i}}=\frac{\partial f}{\partial y_{1}}\frac{% \partial g_{1}}{\partial x_{i}}+\frac{\partial f}{\partial y_{2}}\frac{% \partial g_{2}}{\partial x_{i}}+\frac{\partial f}{\partial y_{3}}\frac{% \partial g_{3}}{\partial x_{i}},\mathbf{\,\,}i=1,2,3 \tag{(1)} \end{equation}% Vz\v{t}ah (1) sa d\'{a} \v{l}ah\v{s}ie zapam\"{a}ta\v{t} v nepresnej\v{s}ej forme. Ak nap\'{\i}\v{s}eme $u=h(\mathbf{\,y\,}),$ $\mathbf{\,y\,}=\mathbf{% \,g}(\mathbf{\,x\,}),$ potom polo\v{z}\'{\i}me $u=f(\mathbf{\,y\,})$ a p% \'{\i}\v{s}eme \begin{equation} \frac{\partial u}{\partial x_{i}}=\frac{\partial u}{\partial y_{1}}\frac{% \partial y_{1}}{\partial x_{i}}+\frac{\partial u}{\partial y_{2}}\frac{% \partial y_{2}}{\partial x_{i}}+\frac{\partial u}{\partial y_{3}}\frac{% \partial y_{3}}{\partial x_{i}}. \tag{(2)} \end{equation}% Posledn\'{y} vz\v{t}ah sa naz\'{y}va \label{6}\emph{re\v{t}azov\'{e} pravidlo.} \end{example} \begin{example} Nech $\mathbf{g}(x,y)=(x^{2}-y^{2}+xy,y^{2}-1)$ a $\mathbf{f}(u,v)=(u+v,2u,% \mathbf{\,}v^{2}).$ Uk\'{a}\v{z}te, \v{z}e $\mathbf{g},\,\mathbf{f}$ s\'{u} diferencovate\v{l}n\'{e}, $\mathbf{f\circ g}$ existuje a vypo\v{c}\'{\i}% tajte deriv\'{a}ciu funkcie $\mathbf{f\circ g}$ v bode $(1,1)$ a) priamo, b) pomocou re\v{t}azov\'{e}ho pravidla. \end{example} \begin{solution} a) $\mathbf{g}(x,y)=(g_{1}(x,y),g_{2}(x,y)),\mathbf{\,}% g_{1}(x,y)=x^{2}-y^{2}+xy,\mathbf{\,}g_{2}(x,y)=y^{2}-1.$ Potom \[ \mathbf{\,}\frac{\partial g_{1}}{\partial x}=2x+y,\frac{\partial g_{1}}{% \partial y}=-2y+x,\mathbf{\,}\frac{\partial g_{2}}{\partial x}=0,\mathbf{\,}% \frac{\partial g_{2}}{\partial y}=2y. \] Ka\v{z}d\'{a} parci\'{a}lna deriv\'{a}cia je evidentne spojit\'{a}, teda $% \mathbf{g\,}\in C^{1}(\mathbf{R}^{2},\mathbf{R}^{2})$ a teda je diferencovate% \v{l}n\'{a}. Podobne $\mathbf{f\,}\in C^{1}(\mathbf{R}^{2},\mathbf{R}^{3}).$ Potom $\mathbf{h}:\mathbf{R}^{2}\longrightarrow \mathbf{R}^{3},\,\mathbf{h}=% \mathbf{f\circ g,}$ preto\v{z}e $\mathbf{g}(\mathbf{R}^{2})\subseteq \mathbf{% R}^{2}.$ Potom \[ (\mathbf{f\circ g})(x,y)=\mathbf{f}% (g_{1}(x,y),g_{2}(x,y))=(g_{1}+g_{2},2g_{1},g_{2}^{2})= \]% \[ =(x^{2}+xy-1,2x^{2}-2y^{2}+2xy,y^{4}-2y^{2}+1)=(h_{1},h_{2},h_{3}). \] Potom% \[ D(\mathbf{f\circ g})_{(x,y)}=D\mathbf{h}_{(x,y)}=\left( \begin{array}{cc} \frac{\partial h_{1}\left( x,y\right) }{\partial x} & \frac{\partial h_{1}\left( x,y\right) }{\partial y} \\ \frac{\partial h_{2}\left( x,y\right) }{\partial x} & \frac{\partial h_{2}\left( x,y\right) }{\partial y} \\ \frac{\partial h_{3}\left( x,y\right) }{\partial x} & \frac{\partial h_{3}\left( x,y\right) }{\partial y}% \end{array}% \right) =\left( \begin{array}{cc} 2x+y & x \\ 4x+2y & -4y+2x \\ 0 & 4y^{3}-4y% \end{array}% \right) \]% a teda \[ D\mathbf{h}_{(1,1)}=\left( \begin{array}{cc} 3 & 1 \\ 6 & -2 \\ 0 & 0% \end{array}% \right) . \]% b) Teraz pou\v{z}ijeme re\v{t}azov\'{e} pravidlo a dostaneme \[ D(\mathbf{f\circ g})_{(1,1)}=D\mathbf{f}_{\left( 1,0\right) }D\mathbf{g}% _{\left( 1,1\right) }==\left( \begin{array}{cc} \frac{\partial f_{1}\left( 1,0\right) }{\partial u} & \frac{\partial f_{1}\left( 1,0\right) }{\partial v} \\ \frac{\partial f_{2}\left( 1,0\right) }{\partial u} & \frac{\partial f_{2}\left( 1,0\right) }{\partial v} \\ \frac{\partial f_{3}\left( 1,0\right) }{\partial u} & \frac{\partial f_{3}\left( 1,0\right) }{\partial v}% \end{array}% \right) \left( \begin{array}{cc} \frac{\partial g_{1}\left( 1,1\right) }{\partial x} & \frac{\partial g_{1}\left( 1,1\right) }{\partial y} \\ \frac{\partial g_{2}\left( 1,1\right) }{\partial x} & \frac{\partial g_{2}\left( 1,1\right) }{\partial y}% \end{array}% \right) = \]% \[ =\left( \begin{array}{cc} 1 & 1 \\ 2 & 0 \\ 0 & 0% \end{array}% \right) \left( \begin{array}{cc} 3 & -1 \\ 0 & 2% \end{array}% \right) =\left( \begin{array}{cc} 3 & 1 \\ 6 & -2 \\ 0 & 0% \end{array}% \right) , \]% teda sme dostali tak\'{y} ist\'{y} v\'{y}sledok ako predt\'{y}m. $\square $ \end{solution} \begin{example} Nech $w=f(x,y,z)$ a predpokladajme, \v{z}e $z=\Phi (x,y).$ Vypo\v{c}\'{\i}% tajte $\frac{\partial w}{\partial x}$ a $\frac{\partial w}{\partial y}.$ $% \mathbf{\,}$ \end{example} \begin{solution} Preto\v{z}e $z=\Phi (x,y),\mathbf{\,}w=f(x,y,\Phi (x,y))=h(x,y)$ a my chceme vypo\v{c}\'{\i}ta\v{t} $\frac{\partial h}{\partial x}$ a $\frac{\partial h}{% \partial x}.$Ak budeme p\'{\i}sa\v{t} $\mathbf{g}(x,y)=(x,y,\Phi (x,y)),$ potom $h(x,y)=(f\mathbf{\circ g})(x,y)$ a pod\v{l}a re\v{t}azov\'{e}ho pravidla \[ Dh_{(x,y)}=Df_{(g_{1},g_{2},g_{3})}D\mathbf{g}_{(x,y)} \]% odkia\v{l} \[ Dh_{\left( x,y\right) }=\left( \frac{\partial f}{\partial x},\frac{\partial f% }{\partial y},\frac{\partial f}{\partial z}\right) \left( \begin{array}{cc} \frac{\partial g_{1}}{\partial x} & \frac{\partial g_{1}}{\partial y} \\ \frac{\partial g_{2}}{\partial x} & \frac{\partial g_{2}}{\partial y} \\ \frac{\partial g_{3}}{\partial x} & \frac{\partial g_{3}}{\partial y}% \end{array}% \right) =\left( \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},% \frac{\partial f}{\partial z}\right) \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \frac{\partial \Phi }{\partial x} & \frac{\partial \Phi }{\partial y}% \end{array}% \right) , \]% odkia\v{l} \[ \frac{\partial f}{\partial x}=\frac{\partial f}{\partial x}+\frac{\partial f% }{\partial z}\frac{\partial \Phi }{\partial x}\text{ \ a \ \ }\frac{\partial f}{\partial y}=\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}% \frac{\partial \Phi }{\partial y}. \]% Tento v\'{y}sledok v\v{s}ak m\^{o}$\mathbf{\,}$\v{z}me dosta\v{t} jednoduch% \v{s}ie pou\v{z}it\'{\i}m verzie re\v{t}azov\'{e}ho pravidla z %TCIMACRO{\hyperref{pr\'{\i}kladu}{}{}{Ma33.tex#3}}% %BeginExpansion \msihyperref{pr\'{\i}kladu}{}{}{Ma33.tex#3}% %EndExpansion . Skuto\v{c}ne, ak nap\'{\i}\v{s}eme \[ w=f(x,y,\mathbf{\,\,}\Phi (x,y)) \]% dost\'{a}vame $\frac{\partial w}{\partial x}=\frac{\partial f}{\partial x}% \frac{\partial x}{\partial x}+\mathbf{\,}\frac{\partial f}{\partial y}\frac{% \partial y}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial \Phi }{% \partial x}=\frac{\partial f}{\partial x}.1+\mathbf{\,}\frac{\partial f}{% \partial y}.0+\frac{\partial f}{\partial z}\frac{\partial \Phi }{\partial x}=% \frac{\partial f}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial \Phi }{\partial x}.\square $ \end{solution} \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{maiindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{maiindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{Ma3.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{Ma3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{Ma32.tex}}% %BeginExpansion \msihyperref{Predch\'{a}dzaj\'{u}ca str\'{a}nka}{}{}{Ma32.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{Ma34.tex}}% %BeginExpansion \msihyperref{Nasleduj\'{u}ca str\'{a}nka}{}{}{Ma34.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{O3.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{O3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Cvi\v{c}enia}{}{}{C3.tex}}% %BeginExpansion \msihyperref{Cvi\v{c}enia}{}{}{C3.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{Glos.tex}}% %BeginExpansion \msihyperref{Index}{}{}{Glos.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \rule{6.5in}{0.04in} \textsl{Matematick\'{a} anal\'{y}za II} \section{Diferencovate\v{l}n\'{e} funkcie} \end{document}