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| -rw-r--r-- | šola/ana1/teor.lyx | 1194 |
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 |