1 files changed, 1193 insertions, 1 deletions

diff --git a/šola/ana1/teor.lyx b/šola/ana1/teor.lyx
index 2b6057d..e44036f 100644
--- a/šola/ana1/teor.lyx
+++ b/šola/ana1/teor.lyx
@@ -9620,7 +9620,20 @@ Enakomerna zveznost
 \begin_inset Formula $f:I\to\mathbb{R}$
 \end_inset
 
- je enakomerno zvezna na 
+ je 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{ez}{enakomerno zvezna}
+\end_layout
+
+\end_inset
+
+ na 
 \begin_inset Formula $I$
 \end_inset
 
@@ -9700,7 +9713,1186 @@ Pri slednji definiciji je
 \begin_inset Formula $\delta$
 \end_inset
 
+
+\end_layout
+
+\begin_layout Theorem*
+Zvezna funkcija na kompaktni množici je enakomerno zvezna.
+\end_layout
+
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $f:K\to\mathbb{R}$
+\end_inset
+
+ zvezna,
+ kjer je 
+\begin_inset Formula $K$
+\end_inset
+
+ kompaktna podmnožica 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni enakomerno zvezna.
+ Zanikajmo definicijo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{ez}{enakomerne zveznosti}
+\end_layout
+
+\end_inset
+
+:
+ 
+\begin_inset Formula $\exists\varepsilon>0\forall\delta>0\exists x_{\delta},y_{\delta}\in I:\left|x_{\delta}-y_{\delta}\right|<\delta\wedge\left|f\left(x_{\delta}\right)-f\left(y_{\delta}\right)\right|\geq\varepsilon$
+\end_inset
+
+.
+ 
+\begin_inset Formula $x,y$
+\end_inset
+
+ sta seveda lahko odvisna od 
+\begin_inset Formula $\delta$
+\end_inset
+
+ in 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ zato v subskriptu pišemo 
+\begin_inset Formula $\delta$
+\end_inset
+
+,
+ ki ji pripadata.
+ Ker smo dejali,
+ da to velja,
+ si oglejmo 
+\begin_inset Formula $\forall n\in\mathbb{N}:\delta_{n}\coloneqq\frac{1}{n}$
+\end_inset
+
+ in pripadajoči zaporedji 
+\begin_inset Formula $\left(x_{1/n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(y_{1/n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ kompaktna,
+ ima zaporedje 
+\begin_inset Formula $\left(x_{1/n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ stekališče v 
+\begin_inset Formula $x\in K$
+\end_inset
+
+,
+ torej obstaja podzaporede 
+\begin_inset Formula $\left(x_{1/n_{k}}\right)_{k\in\mathbb{N}}$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Podobno obstaja podzaporedje 
+\begin_inset Formula $\left(y_{1/n_{k_{l}}}\right)_{l\in\mathbb{N}}$
+\end_inset
+
+,
+ ki konvergira k 
+\begin_inset Formula $y\in K$
+\end_inset
+
+.
+ Pišimo sedaj 
+\begin_inset Formula $x_{l}\coloneqq x_{1/n_{k_{l}}}$
+\end_inset
+
+in 
+\begin_inset Formula $y_{l}\coloneqq y_{1/n_{k_{l}}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Velja torej 
+\begin_inset Formula $x_{l}\to x$
+\end_inset
+
+ in 
+\begin_inset Formula $y_{l}\to y$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $\left|x-y\right|\leq\lim_{l\to\infty}\left(\left|x-x_{l}\right|+\left|x_{l}-y_{l}\right|+\left|y_{l}-y\right|\right)$
+\end_inset
+
 .
+ Levi in desni člen sta v limiti enaka 0 zaradi konvergence zaporedja,
+ srednji pa je manjši od 
+\begin_inset Formula $\frac{1}{j}$
+\end_inset
+
+ zaradi naše predpostavke (PDDRAA),
+ potemtakem je 
+\begin_inset Formula $x=y$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Zato 
+\begin_inset Formula $\lim_{l\to\infty}\left(f\left(x_{l}\right)-f\left(y_{l}\right)\right)=\lim_{l\to\infty}\left[\left(f\left(x_{l}\right)-f\left(x\right)\right)+\left(f\left(x\right)-f\left(y\right)\right)+\left(f\left(y\right)-f\left(y_{l}\right)\right)\right]$
+\end_inset
+
+.
+ Levi in desni člen sta v limiti enaka 0 zaradi konvergence zaporedja in zveznosti 
+\begin_inset Formula $f$
+\end_inset
+
+,
+ srednji pa je tudi 0,
+ ker 
+\begin_inset Formula $x=y$
+\end_inset
+
+,
+ potemtakem 
+\begin_inset Formula $f\left(x_{l}\right)-f\left(y_{l}\right)\to0$
+\end_inset
+
+,
+ kar je v protislovju z 
+\begin_inset Formula $\left|f\left(x_{l}\right)-f\left(y_{l}\right)\right|\geq\varepsilon$
+\end_inset
+
+ za fiksen 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall l\in\mathbb{N}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+,
+ 
+\begin_inset Formula $f$
+\end_inset
+
+ je enakomerno zvezna.
+\end_layout
+
+\begin_layout Corollary*
+En zaprt interval 
+\begin_inset Formula $\frac{1}{x}$
+\end_inset
+
+ bo enakomerno zvezen,
+ 
+\begin_inset Formula $\frac{1}{x}$
+\end_inset
+
+ sama po sebi kot 
+\begin_inset Formula $\left(0,\infty\right)\to\mathbb{R}$
+\end_inset
+
+ pa ni definirana na kompaktni množici.
+ Prav tako 
+\begin_inset Formula $\arcsin$
+\end_inset
+
+ in 
+\begin_inset Formula $x\mapsto\sqrt{x}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Section
+Odvod
+\end_layout
+
+\begin_layout Standard
+Najprej razmislek/ideja.
+ Odvod je hitrost/stopnja,
+ s katero se v danem trenutku neka količina spreminja.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Float figure
+placement document
+alignment document
+wide false
+sideways false
+status open
+
+\begin_layout Plain Layout
+TODO XXX FIXME SKICA S TKZ EUCLID (ali pa —
+ bolje —
+ s čim drugim),
+ glej PS zapiski/ANA1P FMF 2023-12-04.pdf
+\end_layout
+
+\begin_layout Plain Layout
+\begin_inset Caption Standard
+
+\begin_layout Plain Layout
+Skica.
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Radi bi določili naklon sekante,
+ torej naklon premice,
+ določene z 
+\begin_inset Formula $x$
+\end_inset
+
+ in neko bližnjo točko 
+\begin_inset Formula $x+h$
+\end_inset
+
+ na grafu funkcije,
+ ki je odvisen le od 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ ne pa tudi od izbire 
+\begin_inset Formula $h$
+\end_inset
+
+.
+ Bližnjo točko pošljemo proti začetni —
+ 
+\begin_inset Formula $h$
+\end_inset
+
+ pošljemo proti 0.
+ Naklon izračunamo s izrazom 
+\begin_inset Formula $\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Odvod funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $x$
+\end_inset
+
+ označimo 
+\begin_inset Formula $f'\left(x\right)\coloneqq\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+ Če limita obstaja v točki 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ pravimo,
+ da je funkcija odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Pravimo,
+ da je 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva na množici 
+\begin_inset Formula $I\subseteq\mathbb{R}$
+\end_inset
+
+,
+ če je odvedljiva na vsaki 
+\begin_inset Formula $t\in I$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Primeri odvodov preprostih funkcij.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=c,c\in\mathbb{R}$
+\end_inset
+
+ 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{\cancelto{c}{f\left(x+h\right)}-\cancelto{c}{f\left(x\right)}}{h}=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=\lim_{h\to0}\frac{x+h-x}{h}=1$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\left(x\right)=x^{2}$
+\end_inset
+
+ 
+\begin_inset Formula $f'\left(x\right)=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=\lim_{h\to0}\frac{x^{2}+2xh+h^{2}-x^{2}}{h}=\lim_{h\to0}\frac{2xh+h^{2}}{h}=\lim_{h\to0}2x+h=2x$
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+Za poljuben 
+\begin_inset Formula $n\in\mathbb{N}$
+\end_inset
+
+ so funkcije 
+\begin_inset Formula $f\left(x\right)=x^{n}$
+\end_inset
+
+ odvedljive na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $f'\left(x\right)=nx^{n-1}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)=\left(x+h\right)^{n}-x^{n}=\sum_{k=0}^{n}\binom{n}{k}h^{k}x^{n-k}-x^{n}=\cancel{x^{n}}+nhx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k}x^{n-k}\cancel{-x^{n}}}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{nhx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k}x^{n-k}}{h}=\lim_{h\to0}\frac{\cancel{h}\left(nx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k-1}x^{n-k}\right)}{\cancel{h}}=\lim_{h\to0}\left(nx^{n-1}+\sum_{k=2}^{n}\binom{n}{k}h^{k-1}x^{n-k}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=nx^{n-1}+\cancel{\lim_{h\to0}\sum_{k=2}^{n}\binom{n}{k}\cancelto{0}{h^{k-1}}x^{n-k}}=nx^{n-1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Claim*
+\begin_inset Formula $\sin'=\cos$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\cos'=-\sin$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Najprej dokažimo 
+\begin_inset Formula $\sin'=\cos$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\lim_{h\to\infty}\frac{\sin\left(x+h\right)-\sin\left(x\right)=\sin x\cos h+\sin h\cos x-\sin x=\sin x\left(\cos h-1\right)+\sin h\cos x}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\sin x\frac{\left(\cos h-1\right)\left(\cos h+1\right)=\cos^{2}h-1=-\sin^{2}h}{h\left(\cos h+1\right)}+\cos x\frac{\sin h}{h}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\sin x\frac{-\sin h}{h}\cdot\frac{\sin h}{\cos h+1}+\cos x\frac{\sin h}{h}\right)=\lim_{h\to0}\cancelto{1}{\frac{\sin h}{h}}\left(\cos x-\cancel{\sin x\frac{\cancelto{0}{\sin h}}{\cos h+1}}\right)=\cos x
+\]
+
+\end_inset
+
+Sedaj dokažimo še 
+\begin_inset Formula $\cos'=-\sin$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{\cos\left(x+h\right)-\cos\left(x\right)=\cos x\cos h-\sin x\sin h-\cos x=\cos x\left(\cos h-1\right)-\sin x\sin h}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\cos x\frac{\left(\cos h-1\right)\left(\cos h+1\right)=\cos^{2}h-1=-\sin^{2}h}{h\left(\cos h+1\right)}-\sin x\frac{\sin h}{h}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\cos x\frac{-\sin h}{h}\cdot\frac{\sin^{}h}{\cos h+1}-\sin x\frac{\sin h}{h}\right)=\lim_{h\to0}\left(\cancelto{1}{\frac{\sin h}{h}}\left(\cancel{-\cos x\frac{\cancelto{0}{\sin h}}{\cos h+1}}-\sin x\right)\right)=-\sin x
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Fact*
+Od prej vemo 
+\begin_inset Formula $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}=e$
+\end_inset
+
+ (limita zaporedja).
+ Velja tudi 
+\begin_inset Formula $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{x}=e$
+\end_inset
+
+ (funkcijska limita).
+ Ne bomo dokazali.
+\end_layout
+
+\begin_layout Claim*
+Naj bo 
+\begin_inset Formula $a>0$
+\end_inset
+
+ in 
+\begin_inset Formula $f\left(x\right)=a^{x}$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $f'\left(x\right)=a^{x}\ln a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)=a^{x}a^{h}-a^{x}}{h}=\lim_{h\to0}a^{x}\frac{a^{h}-1}{h}=\cdots
+\]
+
+\end_inset
+
+Sedaj pišimo 
+\begin_inset Formula $\frac{1}{z}\coloneqq a^{h}-1$
+\end_inset
+
+.
+ Ulomek 
+\begin_inset Formula $\frac{a^{h}-1}{h}$
+\end_inset
+
+ namreč ni odvisen od 
+\begin_inset Formula $x$
+\end_inset
+
+.
+ Sedaj
+\begin_inset Formula 
+\[
+a^{h}-1=\frac{1}{z}
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+a^{h}=\frac{1}{z}+1
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+h=\log_{a}\left(\frac{1}{z}+1\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+h=\frac{\ln\left(\frac{1}{z}+1\right)}{\ln a}
+\]
+
+\end_inset
+
+Nadaljujmo s prvotnim računom,
+ ločimo primere:
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $a>1,h\searrow0$
+\end_inset
+
+ Potemtakem 
+\begin_inset Formula $a^{h}-1\searrow0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\frac{1}{z}\searrow0$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $z\nearrow\infty$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\cdots=a^{x}\lim_{z\to\infty}\frac{\frac{1}{z}}{\frac{\ln\left(\frac{1}{z}+1\right)}{\ln a}}=a^{x}\lim_{z\to\infty}\frac{\frac{1}{z}\ln a}{\ln\left(\frac{1}{z}+1\right)}=a^{x}\lim_{z\to\infty}\frac{\ln a}{\ln\left(\frac{1}{z}+1\right)^{z}}=a^{x}\lim_{z\to\infty}\frac{\ln a}{\ln e}=a^{x}\ln a
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $a>1,h\nearrow0$
+\end_inset
+
+ Potemtakem 
+\begin_inset Formula $a^{h}-1\nearrow0$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\frac{1}{z}\nearrow0$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $z\searrow-\infty$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\cdots=a^{x}\lim_{z\to-\infty}\frac{\ln a}{\ln\left(\frac{1}{z}+1\right)^{z}}=a^{x}\lim_{z\to-\infty}\frac{\ln a}{\ln\cancelto{e}{\left(\frac{1}{z}+1\right)^{z}}}=a^{x}\ln a
+\]
+
+\end_inset
+
+Kajti 
+\begin_inset Formula $\lim_{x\to\infty}\left(1+\frac{k}{x}\right)^{x}=e^{k}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $a\in(0,1]$
+\end_inset
+
+ Podobno kot zgodaj,
+ bodisi 
+\begin_inset Formula $z\nearrow\infty$
+\end_inset
+
+ bodisi 
+\begin_inset Formula $z\searrow-\infty$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Claim*
+Če je 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva v točki 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ je tam tudi zvezna.
+\end_layout
+
+\begin_layout Proof
+Predpostavimo,
+ da obstaja limita 
+\begin_inset Formula $\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+.
+ Želimo dokazati 
+\begin_inset Formula $f\left(x\right)=\lim_{t\to x}f\left(t\right)$
+\end_inset
+
+.
+ Računajmo:
+\begin_inset Formula 
+\[
+f\left(x\right)=\lim_{t\to x}f\left(t\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0=\lim_{t\to x}f\left(t\right)-f\left(x\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0=\lim_{h\to0}f\left(x+h\right)-f\left(x\right)
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+0=\lim_{h\to0}\left(f\left(x+h\right)-f\left(x\right)\right)=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\cdot h\right)
+\]
+
+\end_inset
+
+Limita obstaja,
+ čim obstajata 
+\begin_inset Formula $\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
+\end_inset
+
+,
+ ki obstaja po predpostavki,
+ in 
+\begin_inset Formula $\lim_{h\to0}h$
+\end_inset
+
+,
+ ki obstaja in ima vrednost 
+\begin_inset Formula $0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\left|x\right|=\sqrt{x^{2}}$
+\end_inset
+
+.
+ Je zvezna,
+ ker je kompozitum zveznih funkcij,
+ toda v 
+\begin_inset Formula $0$
+\end_inset
+
+ ni odvedljiva,
+ kajti 
+\begin_inset Formula $\lim_{h\to0}\frac{f\left(0+h\right)-f\left(0\right)}{h}=\lim_{h\to0}\frac{\left|h\right|-0}{h}=\lim_{h\to0}\sgn h$
+\end_inset
+
+.
+ Limita ne obstaja,
+ ker 
+\begin_inset Formula $-1=\lim_{h\nearrow0}\sgn h\not=\lim_{h\searrow0}\sgn h=1$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bosta 
+\begin_inset Formula $f,g$
+\end_inset
+
+ odvedljivi v 
+\begin_inset Formula $x\in\mathbb{R}$
+\end_inset
+
+.
+ Tedaj so 
+\begin_inset Formula $f+g,f-g,f\cdot g,f/g$
+\end_inset
+
+ (slednja le,
+ če 
+\begin_inset Formula $g\left(x\right)\not=0$
+\end_inset
+
+) in velja 
+\begin_inset Formula $\left(f\pm g\right)'=f'\pm g'$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(fg\right)'=f'g+fg'$
+\end_inset
+
+,
+ 
+\begin_inset Formula $\left(f/g\right)'=\frac{f'g-fg'}{g^{2}}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokažimo vse štiri trditve.
+\end_layout
+
+\begin_deeper
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f+g$
+\end_inset
+
+ Velja 
+\begin_inset Formula $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$
+\end_inset
+
+.
+\begin_inset Formula 
+\[
+\left(f+g\right)'\left(x\right)=\lim_{h\to0}\frac{\left(f+g\right)\left(x+h\right)-\left(f+g\right)\left(x\right)=f\left(x+h\right)+g\left(x+h\right)-f\left(x\right)-g\left(x\right)}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}+\frac{g\left(x+h\right)-g\left(x\right)}{h}\right)=f\left(x\right)'+g\left(x\right)'
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $-f$
+\end_inset
+
+ Naj bo 
+\begin_inset Formula $g=-f$
+\end_inset
+
+.
+ 
+\begin_inset Formula $g'\left(x\right)=\lim_{h\to0}\frac{g\left(x+h\right)-g\left(x\right)}{h}=\lim_{h\to0}\frac{-f\left(x+h\right)+f\left(x\right)}{h}=-\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}=-f\left(x\right)'$
+\end_inset
+
+,
+ zato
+\begin_inset Formula 
+\[
+\left(f-g\right)'\left(x\right)=\left(f+\left(-g\right)\right)'\left(x\right)=f'\left(x\right)+\left(-g\right)'\left(x\right)=f'\left(x\right)-g'\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f\cdot g$
+\end_inset
+
+ Velja 
+\begin_inset Formula $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)$
+\end_inset
+
+.
+ Prištejemo in odštejemo isti izraz (v oglatih oklepajih).
+\begin_inset Formula 
+\[
+\left(fg\right)'\left(x\right)=\lim_{h\to0}\frac{\left(fg\right)\left(x+h\right)-\left(fg\right)\left(x\right)=f\left(x+h\right)g\left(x+h\right)-f\left(x\right)g\left(x\right)+\left[f\left(x\right)g\left(x+h\right)-f\left(x\right)g\left(x+h\right)\right]}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{g\left(x+h\right)\left(f\left(x+h\right)-f\left(x\right)\right)+f\left(x\right)\left(g\left(x+h\right)-g\left(x\right)\right)}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{f\left(x+h\right)-f\left(x\right)}{h}\cancelto{g\left(x\right)}{g\left(x+h\right)}+\lim_{h\to0}\frac{g\left(x+h\right)-g\left(x\right)}{h}f\left(x\right)=f'\left(x\right)g\left(x\right)+g'\left(x\right)f\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Labeling
+\labelwidthstring 00.00.0000
+\begin_inset Formula $f/g$
+\end_inset
+
+ Velja 
+\begin_inset Formula $\left(f/g\right)\left(x\right)=f\left(x\right)/g\left(x\right)$
+\end_inset
+
+.
+ Prištejemo in odštejemo isti izraz (v oglatih oklepajih).
+\begin_inset Formula 
+\[
+\left(f/g\right)'\left(x\right)=\lim_{h\to0}\frac{\left(f/g\right)\left(x+h\right)-\left(f/g\right)\left(x\right)=\frac{f\left(x+h\right)}{g\left(x+h\right)}-\frac{f\left(x\right)}{g\left(x\right)}=\frac{f\left(x+h\right)g\left(x\right)}{g\left(x+h\right)g\left(x\right)}-\frac{f\left(x\right)g\left(x+h\right)}{g\left(x\right)g\left(x+h\right)}=\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)}{g\left(x\right)g\left(x+h\right)}}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)=f\left(x+h\right)g\left(x\right)-f\left(x\right)g\left(x+h\right)+\left[f\left(x\right)g\left(x\right)-f\left(x\right)g\left(x\right)\right]}{hg\left(x\right)g\left(x+h\right)}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\cdot\frac{g\left(x\right)}{g\left(x\right)g\left(x+h\right)}-\frac{g\left(x+h\right)-g\left(x\right)}{h}\cdot\frac{f\left(x\right)}{g\left(x\right)g\left(x+h\right)}\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\left(\left(\frac{1}{g\left(x\right)g\left(x+h\right)}\right)\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}g\left(x\right)-\frac{g\left(x+h\right)-g\left(x\right)}{h}f\left(x\right)\right)\right)=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\frac{1}{g^{2}\left(x\right)}\left(f'\left(x\right)g\left(x\right)-g'\left(x\right)f\left(x\right)\right)=\frac{f'\left(x\right)g\left(x\right)-f\left(x\right)g'\left(x\right)}{g^{2}\left(x\right)}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\end_deeper
+\begin_layout Example*
+\begin_inset Formula $\tan'\left(x\right)=\left(\frac{\sin\left(x\right)}{\cos\left(x\right)}\right)'=\frac{\sin'\left(x\right)\cos\left(x\right)-\sin\left(x\right)\cos'\left(x\right)}{\cos^{2}\left(x\right)}=\frac{\cos^{2}\left(x\right)+\sin^{2}\left(x\right)}{\cos^{2}\left(x\right)}=\cos^{-2}\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+.
+ Tedaj je 
+\begin_inset Formula $g\circ f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g'\left(f\left(x\right)\right)\cdot f'\left(x\right)$
+\end_inset
+
+ (opomba:
+ 
+\begin_inset Formula $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$
+\end_inset
+
+).
+\end_layout
+
+\begin_layout Proof
+Označimo 
+\begin_inset Formula $a\coloneqq f\left(x\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\delta_{h}\coloneqq f\left(x+h\right)-f\left(x\right)$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $f\left(x+h\right)\coloneqq a+\delta\left(h\right)$
+\end_inset
+
+.
+ 
+\begin_inset Formula 
+\[
+\lim_{h\to0}\frac{\left(g\circ f\right)\left(x+h\right)-\left(g\circ f\right)\left(x\right)=g\left(f\left(x+h\right)\right)-g\left(f\left(x\right)\right)=g\left(a+\delta_{h}\right)-g\left(a\right)}{h}=\lim_{h\to0}\frac{g\left(a+\delta_{h}\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{\delta_{h}}{h}=
+\]
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+=\lim_{h\to0}\frac{g\left(a+\delta_{h}\right)-g\left(a\right)}{\delta_{h}}\cdot\frac{f\left(x+h\right)-f\left(x\right)}{h}=\cdots
+\]
+
+\end_inset
+
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ odvedljiva v 
+\begin_inset Formula $x$
+\end_inset
+
+,
+ je v 
+\begin_inset Formula $x$
+\end_inset
+
+ zvezna,
+ zato sledi 
+\begin_inset Formula $h\to0\Rightarrow\delta_{h}\to0$
+\end_inset
+
+,
+ torej
+\begin_inset Formula 
+\[
+\cdots=g'\left(a\right)\cdot f'\left(x\right)=g'\left(f\left(x\right)\right)\cdot f'\left(x\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\varphi\left(x\right)=\sin\left(x^{2}\right)=\left(g\circ f\right)\left(x\right),f\left(x\right)=x^{2},g\left(x\right)=\sin x$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\varphi'\left(x\right)=g'\left(f\left(x\right)\right)f'\left(x\right)=\sin'\left(x^{2}\right)\left(x^{2}\right)'=2x\cos\left(x^{2}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\psi\left(x\right)=\sin^{2}\left(x\right)=\left(g\circ f\right)\left(x\right),f\left(x\right)=\sin,g\left(x\right)=x^{2}$
+\end_inset
+
+ in velja 
+\begin_inset Formula $\psi'\left(x\right)=g'\left(f\left(x\right)\right)f'\left(x\right)=2\sin x\cos x=\sin2x$
+\end_inset
+
+ (sinus dvojnega kota)
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\delta'\left(x\right)=\sin\left(e^{x^{2}}\right)=\sin\left(e^{\left(x^{2}\right)}\right)=\left(g\circ h\circ f\right)\left(x\right),g\left(x\right)=\sin x,h\left(x\right)=e^{x},f\left(x\right)=x^{2}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\delta'\left(x\right)=\cos\left(e^{x^{2}}\right)e^{x^{2}}2x$
+\end_inset
+
+,
+ kajti 
+\begin_inset Formula $\left(e^{x}\right)'=e^{x}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $f:I\subseteq\mathbb{R}\to\mathbb{R}$
+\end_inset
+
+ je zvezno odvedljiva na 
+\begin_inset Formula $I$
+\end_inset
+
+,
+ če je na 
+\begin_inset Formula $I$
+\end_inset
+
+ odvedljiva in je 
+\begin_inset Formula $f'$
+\end_inset
+
+ na 
+\begin_inset Formula $I$
+\end_inset
+
+ zvezna.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\begin{cases}
+x^{2}\sin\frac{1}{x} & ;x\not=0\\
+0 & ;x=0
+\end{cases}$
+\end_inset
+
+ je na 
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+ odvedljiva,
+ a ne zvezno.
+ Odvedljivost na 
+\begin_inset Formula $\mathbb{R}\setminus\left\{ 0\right\} $
+\end_inset
+
+ je očitna,
+ preverimo še odvedljivost v 
+\begin_inset Formula $0$
+\end_inset
+
+:
+\begin_inset Formula 
+\[
+f'\left(0\right)=\lim_{h\to0}\frac{f\left(h\right)-0}{h}=\lim_{h\to0}\cdots\text{NADALJUJEM JUTRI}
+\]
+
+\end_inset
+
+
 \end_layout
 
 \begin_layout Standard