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diff --git a/assignment2.tex b/assignment2.tex new file mode 100644 index 0000000..fa70c40 --- /dev/null +++ b/assignment2.tex @@ -0,0 +1,112 @@ +\documentclass[10pt,a4paper]{article} + +\usepackage[margin=2cm]{geometry} + +\usepackage{enumitem} +\setenumerate[1]{label=\arabic*.} +\setenumerate[2]{label=(\alph*)} + +% textcomp package is not available everywhere, and we only need the Copyright symbol +% taken from http://tex.stackexchange.com/a/1677/23992 +\DeclareTextCommandDefault{\textregistered}{\textcircled{\check@mathfonts\fontsize\sf@size\z@\math@fontsfalse\selectfont R}} + +\usepackage{fancyhdr} +\renewcommand{\headrulewidth}{0pt} +\renewcommand{\footrulewidth}{0pt} +\fancyhead{} +\fancyfoot[C]{Copyright {\textcopyright} 2015 Camil Staps} +\pagestyle{fancy} + +\usepackage{amsmath} +\usepackage{amsfonts} + +\parindent0pt + +\title{Calculus en Kansrekenen - assignment 2} +\author{Camil Staps\\\small{s4498062, group Bram}} + +\begin{document} + +\maketitle +\thispagestyle{fancy} + +\begin{enumerate} + \item \begin{enumerate} + \item $$\lim\limits_{x\to-\infty}\frac{x^3+2x^2+2}{3x^3+x+4} = \lim\limits_{x\to-\infty}\frac{1+\frac2x+\frac2{x^3}}{3+\frac1x+\frac4{x^3}} = \frac13.$$ + \item $$\lim\limits_{x\to\infty}\frac{3x^2+8}{x+1}=\frac{3x+\frac8x}{1+\frac1x}=\lim\limits_{x\to\infty}3x+\frac8x=\lim\limits_{x\to\infty}3x=\infty.$$ + \item $$\lim\limits_{x\to\infty}\frac{2x+1}{x^2+x}=\lim\limits_{x\to\infty}\frac{\frac2x+\frac1{x^2}}{1+\frac1x}=\frac01=0.$$ + \item \begin{align*} + & \lim\limits_{x\to a}\frac{x^n-a^n}{x-a}\\ + = & \lim\limits_{a+h\to a}\frac{(a+h)^n-a^n}{a+h-a}\qquad\text{with $h=x-a$}\\ + = & \lim\limits_{h\to0}\frac{(a+h)^n - a^n}{h}\\ + = & \left(a^n\right)'\\ + = & n\cdot a^{n-1}. + \end{align*} + \end{enumerate} + + \item \begin{enumerate} + \item $$f'(2) = \lim\limits_{h\to0}\frac{f(2+h)-f(2)}{h}=\lim\limits_{h\to0}\frac0h=0.$$ + \item $$f'(x) = \lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} = \lim\limits_{h\to0}\frac{2(x+h)+3-(2x+3)}{h} = \lim\limits_{h\to0}2 = 2.$$ + \item $$f'(3) = \lim\limits_{h\to0}\frac{f(3+h)-f(3)}{h} = \lim\limits_{h\to0}\frac{h^2+6h+9+6+2h+1-(9+6+1)}{h} = \lim\limits_{h\to0}h+8=8.$$ + \item \begin{align*} + f'(x) &= \lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}\\ + &= \lim\limits_{h\to0}\frac{\frac{5(x+h)-7}{4(x+h)+3} - \frac{5x-7}{4x+3}}{h}\\ + &= \lim\limits_{h\to0}\frac{\frac{43h}{(4x+4h+3)(4x+3)}}{h}\\ + &= \lim\limits_{h\to0}\frac{43}{(4x+4h+3)(4x+3)}\\ + &= \frac{43}{(4x+3)^2}. + \end{align*} + \end{enumerate} + + \item \begin{enumerate} + \item We rewrite this to something of the form $y=ax+b$: + + \begin{align*} + 4y - x + 2(1-\ln2) &= 0\\ + 4y &= x - 2(1-\ln2)\\ + y &= \frac14x - \frac12(1-\ln2) + \end{align*} + + For a line $y=ax+b$, $a$ is the slope, so in this case we have the slope $\frac14$. + + \item We equate $f(x)$ and $y$ for $x=2$: + + \begin{align*} + \frac14\cdot2 - \frac12(1-\ln2) &= a\ln2 \\ + \frac12\ln2 &= a\ln2 \\ + a &= \frac12. + \end{align*} + + \item For the slope we take the derivative of $f$, i.e. $f'(x) = \frac ax = \frac12a$. + + Then as intercept we take $f(2)-2\cdot f'(2) = a\ln2 - 2\cdot\frac a2 = a\ln2-a$. + + The tangent line is then $y=\frac12ax+a\ln2-a$. + \end{enumerate} + + \item \begin{enumerate} + \item $f'(x) = 4x^3-6x^2.$ + \item $f'(x) = \frac{(x-7)(2x)-(x^2+5)(1)}{(x-7)^2} = \frac{x+2}{x-7}.$ + \item \begin{align*} + f'(x) &= \left(\sin{\sqrt x}\cdot\sin{\sqrt x}\right) \\ + &= 2\cdot\sin\sqrt x\cdot\left(\sin\sqrt x\right)' \\ + &= 2\cdot\sin\sqrt x\cdot\cos\sqrt x\cdot\frac12\cdot x^{-\frac12} \\ + &= \frac12\sin\left(2\sqrt x\right)\cdot\frac1{\sqrt x} \\ + &= \frac{\sin\left(2\sqrt x\right)}{2\sqrt x}. + \end{align*} + \item %todo + \item $f'(x) = \left(e^{\tan x}\right)' = e(\tan x)\cdot\tan'x = \frac{e(\tan x)}{\cos^2x}$. + \item $f'(x) = -\frac1{\cos x}\cdot-\sin x=\frac{\sin x}{\tan x}$. + \item \begin{align*} + f'(x) &= \frac1{\sqrt{1-(1-2x)^2}}\cdot-2 \\ + &= -2\cdot\frac1{\sqrt{4x^2+4x}} \\ + &= -2\cdot\frac1{2\sqrt{x^2+x}} \\ + &= -\left(x^2+x\right)^{-\frac12}. + \end{align*} + \item %todo + \end{enumerate} + + \item % todo +\end{enumerate} + +\end{document} + |