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\title{Math Technology Lab - Day 1}
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\author{Aalliyah Celestine}
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\begin{document}
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\maketitle
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\section{Introduction}
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In this document we discuss the definition of derivatives as limits of the difference quotient, and illustrate it with some examples.
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\section{The definition of derivative} \label{sec2}
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Let $f$ be a function of a real variable whose formula contains the interval $(a,b)$, and let $x$ be a real number in $(a,b)$. We then define
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\begin{equation} \label{defder}
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f'(x)=\lim_{h\rightarrow h}\frac{f(x+h)-f(x)}{h},\end{equation}
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provided the limit exist.
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\begin{example} \label{dx2}
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Suppose $f(x)=x^2$. We will show that $f'(x)=2x$. Using the definition (\ref{defder}), we find
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\begin{eqnarray*}
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f'(x)&=&\lim_{h\rightarrow 0}\frac{(x+h)^2-x^2}{h}\\[2ex]&=& \lim_{h\rightarrow 0}\frac{x^2+2xh+h^2-x^2}{h}\\[2ex]&=&\lim_{h\rightarrow 0}\frac{2xh+h^2}{h}\\[2ex]&=&\lim_{h\rightarrow 0}\frac{h(2x+h)}{h}\\[2ex]&=&\lim_{h\rightarrow 0}(2x+h)\\[2ex]&=&2x
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\end{eqnarray*}
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\end{example}
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The calculation in example ( \ref{dx2}) is a special case of the general power rule: if $f(x)=x^n$, then
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$$f'(x)=nx^{n-1}$$
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\section{Linearization}
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As an application of the derivative that we studied in section (\ref{sec2}), we now use the tangent line to a function as a way to approximate the function near a given point.
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Some functions are easy to compute at a few points, but not so easy at most points.
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\begin{example}
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If $f(x)=\sqrt{x}$, then $f(4)=\sqrt{4}=2$ and $f(16)=\sqrt{16}=4$, but what is $f(7)$?
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\end{example}
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\begin{example}
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If $f(x)=\ln x$, then $f(1)=\ln 1=0$ and $f(e)=\ln e=1$, but what is $f(2)$?
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\end{example}
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\begin{example}
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If $f(x)=\sin x$, then $f(0)=\sin(0)=0$, $f(\pi/2)=\sin(\pi/2)=1$, and $f(\pi/6)=\sin(\pi/6)=(1/2)$, but what is $f(1)$?
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\end{example}
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The simplest type of functions are the linear functions, that is the functions form
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\begin{eqnarray*}
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f(x)=mx+b,
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\end{eqnarray*}
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whose graph is a straight line. These functions are easy to compute at all points.
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In this section, we will study a procedure called linearization that allows us to find approximate values of a function $f(x)$ "near" a given point $x=a$ using only the values of $f(x)$ and its derivative $f'(x)$ at $x=a$.
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That means that if $x=a$ is a point where $f(x)$ and $f'(x)$ are easy to compute, then we can also easily compute $f(x)$ "near" that point.
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Of course saying "near" a point is not a precise description. We will not give a precise, formal definition of this concept, but informally, we will take that to mean that the approximation will get better and better as the point $x$ gets closer and closer to $a$.
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The idea is simple: we approximate a given function $f(x)$ near $x=a$ with its own tangent line at that point. The formula for a straight line $y=mx+b$ is easy to to compute at all points, and we know from looking at a graph that the tangent line to the graph of $f(x)$ at $x=a$ remains quite close to the graph of $f(x)$ as long as we do not go too far from $a$.
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Recall that the equation of the tangent line at $x=a$ is
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\begin{eqnarray*}
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y-f(a)=f'(a)(x-a).
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\end{eqnarray*}
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If we solve this for $y$, we get a linear function: $y=f(a)+f'(a)(x-a)$. We will denote this function by $L_{a}(x)$. So the linearization of $f(x)$ at $x=a$ is
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\begin{equation} \label{lindef}
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L_{a}(x)=f(a)+f'(a)(x-a)
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\end{equation}
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\begin{exercise}
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Let $f(x)=\sqrt{x}$, and $a=25$. Find the linearization $L_{a}(x)$ given by (\ref{lindef}), then use it to estimate $\sqrt{27}$.
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$f(25)=\sqrt{25}=5$.
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\end{exercise}
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\begin{eqnarray*}
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L_a(27)&=&f(25)+f'(25)(27-25)\\[1ex]&=&5+f'(25)(27-25)\\[1ex]&=&5+f'(25)(2)\\[1ex]&=&5+\frac{1}{2\sqrt{25}}(2)\\[1ex]&=&5+\frac{1}{10}(2)\\[1ex]&=&5+\frac{1}{5}\\[1ex]&=&5.2
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\end{eqnarray*}
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\end{document}
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