Merge branch 'Exercise-7' into 'master'
Add exercise 7 and 8 See merge request oysteikt/ma0001-matematikk!1
This commit is contained in:
commit
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*.out
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*.synctex.gz
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*.toc
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*.listing
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\documentclass{article}
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\author{Øystein Tveit}
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\title{MA0001 Øving 2}
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\newcommand{\theTitle}{MA0001 Øving 2}
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\newcommand{\theAuthor}{Øystein Tveit}
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\input{../lib/lib.tex}
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\documentclass{article}
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\newcommand{\theTitle}{MA0001 Øving 3}
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\newcommand{\theAuthor}{Øystein Tveit}
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\input{../lib/lib.tex}
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\begin{document}
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\documentclass{article}
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\newcommand{\theTitle}{MA0001 Øving 4}
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\newcommand{\theAuthor}{Øystein Tveit}
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\input{../lib/lib.tex}
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\usepackage{amssymb}
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\documentclass{article}
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\newcommand{\theTitle}{MA0001 Øving 5}
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\newcommand{\theAuthor}{Øystein Tveit}
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\input{../lib/lib.tex}
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\usepackage{amssymb}
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\documentclass{article}
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\newcommand{\theTitle}{MA0001 Øving 6}
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\newcommand{\theAuthor}{Øystein Tveit}
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\input{../lib/lib.tex}
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\begin{document}
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\documentclass{article}
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\newcommand{\theTitle}{MA0001 Øving 7}
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\newcommand{\theAuthor}{Øystein Tveit}
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\input{../lib/lib.tex}
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\usepackage{amssymb}
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\begin{document}
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\thispagestyle{plain}
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\tittel
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\tableofcontents
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\newpage
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\section{Forberedende oppgaver}
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\begin{oppgaver}
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\oppg
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\input{tasks/1.tex}
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\end{oppgaver}
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\newpage
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\section{Innleveringsoppgaver}
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\begin{oppgaver}
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\setoppg{1}
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\oppg
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\input{tasks/2.tex}
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\oppg
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\input{tasks/3.tex}
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\oppg
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\input{tasks/4.tex}
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\oppg
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\input{tasks/5.tex}
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\end{oppgaver}
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\end{document}
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\begin{deloppgaver}
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\delo
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\[f'(x) = 3\]
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\delo
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\[f^{-1}(x) = \frac{x}{3} - 1\]
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\delo
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\[\frac{d}{dx} f^{-1}(x) = \frac{1}{3} \]
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\end{deloppgaver}
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\[ f(x) = x^2 \]
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\begin{align*}
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\frac{d}{dx} f(x) &= \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} \\[2ex]
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&= \lim_{\Delta x \to 0} \frac{(x+\Delta x)^2 - x^2}{\Delta x} \\[2ex]
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&= \lim_{\Delta x \to 0} \frac{x^2 + 2x \Delta x + {\Delta x}^2 - x^2}{\Delta x} \\[2ex]
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&= \lim_{\Delta x \to 0} \frac{2x \Delta x + {\Delta x}^2}{\Delta x} \\[2ex]
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&= \lim_{\Delta x \to 0} \frac{2x + \Delta x}{1} \\[2ex]
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&= \lim_{\Delta x \to 0} \frac{2x + \Delta x}{1} \\[2ex]
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&= 2x + 0 \\[2ex]
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&= 2x \\[2ex]
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\end{align*}
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\[ h(x) = -\frac{1}{4}x^4 + \frac{2}{3}x^3 + 2x^2 - \sqrt{2}x + \pi^e \]
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\begin{align*}
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h'(x) &= 4 \cdot -\frac{1}{4}x^{4-1} + 3 \cdot \frac{2}{3}x^{3-1} + 2\cdot2x^{2-1} - 1 \cdot \sqrt{2}x^{1-1} + 0 \cdot \pi^e x^{0-1} \\
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&= -x^{3} + 2x^{2} + 4x - \sqrt{2} \\
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\end{align*}
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\begin{align*}
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\left(x^{1/3}-x\right) \cdot \left(x^{2/3}+x^2\right)
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&= x^{1/3} \cdot x^{2/3} + x^{1/3} \cdot x^2 - x\cdot x^{2/3} - x \cdot x^{2} \\
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&= x + x^{7/3} - x^{5/3} - x^3 \\
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\end{align*}
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\begin{align*}
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\frac{d}{dx} \left( x + x^{7/3} - x^{5/3} - x^3 \right)
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&= 1 + \frac{7}{3}x^{4/3} - \frac{5}{3}x^{2/3} - 3x^2 \\
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&= - 3x^2 - \frac{5}{3}x^{2/3} + \frac{7}{3}x^{4/3} + 1
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\end{align*}
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\[ f(x): \mathbb{R} \to \mathbb{R} = \begin{cases}
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\sqrt{x},\qquad &x \ge 0 \\
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x, &x < 0
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\end{cases} \]
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Ettersom \[\sqrt{0} = 0\] og \[\lim_{x\to 0}x = 0\] er funksjonen kontinuerlig.
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Den deriverte av den første delen av funksjonen blir
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\begin{align*}
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\frac{d}{dx} \sqrt{x} &= \frac{1}{2} x^{-\frac{1}{2}} \\[2ex]
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&= \frac{1}{2\sqrt{x}}
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\end{align*}
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Ved $x=0$ blir \[\frac{1}{2\sqrt{x}} = \frac{1}{0}\] som er udefinert, og dermed er ikke den funksjonen deriverbar for denne $x$-verdien
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\documentclass{article}
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\newcommand{\theTitle}{Øystein Tveit}
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\newcommand{\theAuthor}{MA0001 Øving 8}
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\input{../lib/lib.tex}
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\input{../lib/lst.tex}
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\begin{document}
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\thispagestyle{plain}
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\tittel
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\tableofcontents
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\newpage
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\section{Forberedende oppgaver}
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\begin{oppgaver}
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\oppg
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\input{tasks/1.tex}
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\end{oppgaver}
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\newpage
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\section{Innleveringsoppgaver}
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\begin{oppgaver}
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\setoppg{1}
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\oppg
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\input{tasks/2.tex}
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\oppg
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\input{tasks/3.tex}
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\oppg
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\input{tasks/4.tex}
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\oppg
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\input{tasks/5.tex}
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\end{oppgaver}
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\end{document}
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from math import e
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def f(x): return e ** x - 2
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def df(x): return e ** x
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def Newton(a, f, df):
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return a - f(a)/df(a)
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def main():
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i = 0
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a = [5]
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while True:
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a.append(Newton(a[i], f, df))
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if abs(a[i] - a[i-1]) < 0.001:
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print(f'Iterasjoner: {i-1}, a: {a[i-1]}')
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break
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i += 1
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main()
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\[f(x) = \frac{1}{2}x^3 -7x +22\]
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\[f'(x) = \frac{3}{2}x^2 -7\]
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\[f''(x) = 3x \]
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\[f'''(x) = 3 \]
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\[f^{(100)} = 0\]
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\begin{enumerate}[label=\arabic*.]
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\item \[ f(x) = ln\left(\frac{1}{x^2}\right) \]
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La $u = \frac{1}{x^2}$
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Jeg bruker kjerneregelen:
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\begin{align*}
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\frac{dy}{dx} &= \frac{dy}{du} \cdot \frac{du}{dx} \\
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&= \frac{d}{du} ln(u) \cdot \frac{d}{dx} \frac{1}{x^2} \\
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&= \frac{1}{u} -2 \frac{1}{x^3} \\
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&= \frac{1}{\frac{1}{x^2}} -2 \frac{1}{x^3} \\
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&= x^2 -2 \frac{1}{x^3}
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\end{align*}
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\item \[g(x) = \frac{1 + \sin x}{1 + e^x + x^2}\]
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Jeg bruker kvotientregelen:
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$u = 1 + \sin x$
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$v = 1 + e^x + x^2$
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\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{ \frac{d}{dx}(u) \cdot v - u \cdot \frac{d}{dx}(v)}{v^2} \]
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\begin{align*}
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\frac{d}{dx} u &= \frac{d}{dx} 1 + \sin x \\
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&= \cos x
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\end{align*}
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\begin{align*}
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\frac{d}{dx} v &= \frac{d}{dx} 1 + e^x + x^2 \\
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&= e^x + 2x
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\end{align*}
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\begin{align*}
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\frac{dg}{dx} &= \frac{(\cos x)(1+e^x+x^2) - (1+\sin x)(e^x + 2x)}{\left(1 + e^x + x^2\right)^2}
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\end{align*}
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\item \[h(x) = \sqrt{1 + \sqrt{x}}\]
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$u = 1 + \sqrt{x}$
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\begin{align*}
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\frac{dy}{dx} &= \frac{dy}{dv} \cdot \frac{du}{dx} \\
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&= \frac{d}{du} \sqrt{u} \cdot \frac{d}{dx} \left( 1 + \sqrt{x} \right) \\
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&= \frac{1}{2\sqrt{u}} \cdot \frac{1}{2\sqrt{x}} \\
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&= \frac{1}{2\sqrt{1+\sqrt{x}}} \cdot \frac{1}{2\sqrt{x}} \\
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&= \frac{1}{4\sqrt{1+\sqrt{x}} \sqrt{x}}
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\end{align*}
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\end{enumerate}
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\[f(t) = \frac{t^2 -1}{t+1} + 6t^{1/3} + \sqrt{\sin t} + 4^t\]
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Jeg deriverer funksjonen ledd for ledd
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Ledd 1:
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\begin{align*}
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\frac{d}{dt} \frac{t^2-1}{t+1} &= \frac{d}{dt}\frac{(t+1)(t-1)}{t+1} \\
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&= \frac{d}{dt} t-1 \\
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&= 1
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\end{align*}
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Ledd 2:
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\begin{align*}
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\frac{d}{dt} 6t^{1/3} &= \frac{1}{3} \cdot 6t^{(1/3 - 1)} \\
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&= 2t^{-2/3}\\
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&= \frac{2}{\sqrt[3]{t^2}}
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\end{align*}
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Ledd 3:
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\[ \frac{d}{dt} \sqrt{\sin t} \]
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$u = \sin t$
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\begin{align*}
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\frac{dy}{dt} &= \frac{dy}{du} \cdot \frac{du}{dt} \\
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&= \frac{d}{du} \sqrt{u} \cdot \frac{d}{dt} \sin t \\
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&= \frac{1}{2\sqrt{u}} \cdot \cos t \\
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&= \frac{\cos t}{2\sqrt{\sin t}}
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\end{align*}
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Ledd 4:
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\begin{align*}
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\frac{d}{dt} 4^t = 4^t ln(t) \\
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\end{align*}
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\[\frac{df}{dx} = 1 + \frac{2}{\sqrt[3]{t^2}} + \frac{\cos t}{2\sqrt{\sin t}} + 4^t ln(t)\]
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Her har jeg brukt python for å løse oppgaven.
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\pythonBlock{./scripts/4.py}
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Output:
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\fbox{Iterasjoner: 7, a: 0.6932882713164431}
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\[f(y) = a^y \]
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\[f'(y) = a^y \ln a\]
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\[f^{-1}(y) = \log_{a} y\]
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\begin{align*}
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\frac{d}{dy}f^{-1}(y) &= \frac{1}{f' \circ f^{-1}(y)} \\[2ex]
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&= \frac{1}{a^{\log_{a} y} \ln a} \\[2ex]
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&= \frac{1}{y \ln a} \\
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\end{align*}
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Gitt at vi vet at $\frac{d}{dx} ln(x) = \frac{1}{x}$, så kan vi også løse det på følgende måte
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\begin{align*}
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\frac{d}{dy} \log_{a} y &= \frac{d}{dy} \frac{\ln y}{\ln a} \\[2ex]
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&= \frac{1}{\ln a} \cdot \frac{d}{dy} \ln y \\[2ex]
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&= \frac{1}{\ln a} \cdot \frac{1}{y} \\[2ex]
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&= \frac{1}{y \ln a}
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\end{align*}
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\definecolor{ntnublue}{RGB}{0,80,158}
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\input{../lib/geometry.tex}
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\input{../lib/header.tex}
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\input{../lib/geometry.tex}
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\input{../lib/math.tex}
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\usetikzlibrary{angles, quotes}
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\pgfplotsset{compat=newest}
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\author{Øystein Tveit}
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\title{MA0001 Øving 6}
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\author{\theAuthor}
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\title{\theTitle}
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\input{../lib/titling.tex}
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\usepackage{listings}
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\definecolor{background}{RGB}{39, 40, 34}
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\definecolor{string}{RGB}{230, 219, 116}
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\definecolor{comment}{RGB}{117, 113, 94}
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\definecolor{normal}{RGB}{248, 248, 242}
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\definecolor{identifier}{RGB}{166, 226, 46}
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\lstdefinestyle{monokaiPython}{
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frame=none,
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language=python, % choose the language of the code
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numbers=left, % where to put the line-numbers
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stepnumber=1, % the step between two line-numbers.
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numbersep=5pt, % how far the line-numbers are from the code
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numberstyle=\color{white}\ttfamily,
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backgroundcolor=\color{background}, % choose the background color. You must add \usepackage{color}
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showspaces=false, % show spaces adding particular underscores
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showstringspaces=false, % underline spaces within strings
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showtabs=false, % show tabs within strings adding particular underscores
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tabsize=2, % sets default tabsize to 2 spaces
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captionpos=b, % sets the caption-position to bottom
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breaklines=true, % sets automatic line breaking
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breakatwhitespace=true, % sets if automatic breaks should only happen at whitespace
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title=\color{white}\lstname, % show the filename of files included with \lstinputlisting;
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basicstyle=\color{normal}\ttfamily, % sets font style for the code
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keywordstyle=\color{magenta}\ttfamily, % sets color for keywords
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stringstyle=\color{string}\ttfamily, % sets color for strings
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commentstyle=\color{comment}\ttfamily, % sets color for comments
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emph={format_string, eff_ana_bf, permute, eff_ana_btr},
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emphstyle=\color{identifier}\ttfamily
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}
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\usepackage{framed}
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\definecolor{shadecolor}{named}{background}
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\newcommand{\pythonBlock}[1]
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{
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\begin{shaded}
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\lstinputlisting[style=monokaiPython]{#1}
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\end{shaded}
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}
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