\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.4in,right=0.5in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue} \Large Lecture Notes} \chead{\color{black} \LARGE Definition of the Derivative} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ Hidegkuti, Powell, 2014} \rfoot{\small Last revised: September 24, 2015} \textwidth 7.4in \textheight 9.3in \setlength{\headheight}{30pt} \setlength{\parindent}{0in} \begin{document} The derivative of a function of a real variable expresses the rate of change of a quantity (a function value or dependent variable) which is determined by another quantity (the independent variable). Derivatives are a fundamental tool of calculus. For example, the derivative of the location function of a moving object with respect to time is the object's velocity: it measures how quickly the position of the object changes when time is advanced. \bigskip Suppose that $f$ is a function. \ Let $a$ be a fixed number in the domain of $f$.\bigskip The expression $\dfrac{f\left( a+h\right) -f\left( a\right) }{h}$ is called the \textbf{difference quotient}. \ The difference quotient has a geometric meaning: \ it is the slope of the secant line connecting two points on the graph of $f$: \ \ $A\left( a,f\left( a\right) \right) $ and $B\left( a+h,f\left( a+h\right) \right) .$ \ The difference quotient also has an interpretation in physics: if the function $f$ is location function, then the difference quotient expresses the average velocity between times $% t_{1}=a $ and $t_{2}=a+h$.\bigskip \bigskip The \textbf{derivative} of $f$, at the number $a,$ denoted by $f^{\prime }\left( a\right) $, is defined as the limit of the difference quotient:% \begin{equation*} f^{\prime }\left( a\right) =\lim\limits_{h\rightarrow 0}\dfrac{f\left( a+h\right) -f\left( a\right) }{h} \end{equation*}% The derivative of $f$ at $a$ is only defined if the limit shown above exists and is finite.\bigskip \bigskip The derivative also has a geometric meaning: \ $f^{\prime }\left( a\right) $ is the slope of the tangent line drawn to the graph of $f$ at pont $\left( a,f\left( a\right) \right) $. \ The difference quotient also has an interpretation in physics: if the function $f$ is location function, then $% f^{\prime }\left( a\right) $ expresses the instantaneous velocity at $t=a$% .\bigskip \bigskip Given a function $f$, if we evaluate $f^{\prime }\left( x\right) $ for all $% x,$ we obtain a new function, called the derivative (or first derivative) of $f$.\bigskip \bigskip Differentiate each of the following by evaluating the limit of the difference quotient.% %TCIMACRO{\TeXButton{3 col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion \begin{enumerate} \item $f\left( x\right) =x^{2}-3x\bigskip $ \item $f\left( x\right) =x^{3}\bigskip $ \item $f\left( x\right) =\sqrt{2x-1}\bigskip $ \item $f\left( x\right) =\dfrac{1}{x^{2}-1}\bigskip $ \item $f\left( x\right) =\sqrt{1-x^{2}}$ \end{enumerate} %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \pagebreak \begin{center} {\LARGE Solutions \bigskip } \end{center} \begin{enumerate} \item $f\left( x\right) =x^{2}-3x$% \begin{eqnarray*} f^{\prime }\left( x\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right) -f\left( x\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\left( \left( x+h\right) ^{2}-3\left( x+h\right) \right) -\left( x^{2}-3x\right) }{h% } \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{\left( x^{2}+2xh+h^{2}-3x-3h\right) -\left( x^{2}-3x\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{% x^{2}+2xh+h^{2}-3x-3h-x^{2}+3x}{h} \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{2xh+h^{2}-3h}{h}=\lim\limits_{h% \rightarrow 0}\dfrac{h\left( 2x+h-3\right) }{h}=\lim\limits_{h\rightarrow 0}\left( 2x+h-3\right) =2x-3 \end{eqnarray*} \item $f\left( x\right) =x^{3}$% \begin{eqnarray*} f^{\prime }\left( x\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right) -f\left( x\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\left( x+h\right) ^{3}-x^{3}}{h}=\lim\limits_{h\rightarrow 0}\dfrac{% x^{3}+3x^{2}h+3xh^{2}+h^{3}-x^{3}}{h} \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{3x^{2}h+3xh^{2}+h^{3}}{h}% =\lim\limits_{h\rightarrow 0}\dfrac{h\left( 3x^{2}+3xh+h^{2}\right) }{h}% =\lim\limits_{h\rightarrow 0}3x^{2}+3xh+h^{2}=3x^{2} \end{eqnarray*} \item $f\left( x\right) =\sqrt{2x-1}$% \begin{eqnarray*} f^{\prime }\left( x\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right) -f\left( x\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\sqrt{% 2\left( x+h\right) -1}-\sqrt{2x-1}}{h} \\ &=&\lim\limits_{h\rightarrow 0}\left( \dfrac{\sqrt{2x+2h-1}-\sqrt{2x-1}}{h}% \cdot \dfrac{\sqrt{2x+2h-1}+\sqrt{2x-1}}{\sqrt{2x+2h-1}+\sqrt{2x-1}}\right) \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{\left( 2x+2h-1\right) -\left( 2x-1\right) }{h\left( \sqrt{2x+2h-1}+\sqrt{2x-1}\right) }=\lim\limits_{h% \rightarrow 0}\dfrac{2x+2h-1-2x+1}{h\left( \sqrt{2x+2h-1}+\sqrt{2x-1}\right) } \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{2h}{h\left( \sqrt{2x+2h-1}+\sqrt{2x-1}% \right) }=\lim\limits_{h\rightarrow 0}\dfrac{2}{\sqrt{2x+2h-1}+\sqrt{2x-1}}=% \dfrac{2}{2\sqrt{2x-1}}=\dfrac{1}{\sqrt{2x-1}} \end{eqnarray*} \item $f\left( x\right) =\dfrac{1}{x^{2}-1}$% \begin{eqnarray*} f^{\prime }\left( x\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right) -f\left( x\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\dfrac{1% }{\left( x+h\right) ^{2}-1}-\dfrac{1}{x^{2}-1}}{h} \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{\dfrac{x^{2}-1}{\left( \left( x+h\right) ^{2}-1\right) \left( x^{2}-1\right) }-\dfrac{\left( x+h\right) ^{2}-1}{\left( x^{2}-1\right) \left( \left( x+h\right) ^{2}-1\right) }}{h}% =\lim\limits_{h\rightarrow 0}\left( \dfrac{x^{2}-1-\left( \left( x+h\right) ^{2}-1\right) }{\left( \left( x+h\right) ^{2}-1\right) \left( x^{2}-1\right) }\cdot \dfrac{1}{h}\right) \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{x^{2}-1-\left( x^{2}+2xh+h^{2}-1\right) }{h\left( \left( x+h\right) ^{2}-1\right) \left( x^{2}-1\right) }=\lim\limits_{h\rightarrow 0}\left( \dfrac{% x^{2}-1-x^{2}-2xh-h^{2}+1}{h\left( \left( x+h\right) ^{2}-1\right) \left( x^{2}-1\right) }\right) \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{-2xh-h^{2}}{h\left( \left( x+h\right) ^{2}-1\right) \left( x^{2}-1\right) }=\lim\limits_{h\rightarrow 0}\dfrac{% -h\left( 2x+h\right) }{h\left( \left( x+h\right) ^{2}-1\right) \left( x^{2}-1\right) }=\lim\limits_{h\rightarrow 0}\dfrac{-\left( 2x+h\right) }{% \left( \left( x+h\right) ^{2}-1\right) \left( x^{2}-1\right) } \\ &=&\dfrac{-2x}{\left( x^{2}-1\right) \left( x^{2}-1\right) }=\dfrac{-2x}{% \left( x^{2}-1\right) ^{2}} \end{eqnarray*} \item $f\left( x\right) =\sqrt{1-x^{2}}$% \begin{eqnarray*} f^{\prime }\left( x\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right) -f\left( x\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\sqrt{% 1-\left( x+h\right) ^{2}}-\sqrt{1-x^{2}}}{h} \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{\sqrt{1-\left( x+h\right) ^{2}}-\sqrt{% 1-x^{2}}}{h}\cdot \dfrac{\sqrt{1-\left( x+h\right) ^{2}}+\sqrt{1-x^{2}}}{% \sqrt{1-\left( x+h\right) ^{2}}+\sqrt{1-x^{2}}} \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{1-\left( x+h\right) ^{2}-\left( 1-x^{2}\right) }{h\left( \sqrt{1-\left( x+h\right) ^{2}}+\sqrt{1-x^{2}}% \right) }=\lim\limits_{h\rightarrow 0}\dfrac{1-x^{2}-h^{2}-2xh-1+x^{2}}{% h\left( \sqrt{1-\left( x+h\right) ^{2}}+\sqrt{1-x^{2}}\right) } \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{-h^{2}-2xh}{h\left( \sqrt{1-\left( x+h\right) ^{2}}+\sqrt{1-x^{2}}\right) }=\lim\limits_{h\rightarrow 0}\dfrac{-% \NEG{h}\left( h+2x\right) }{\NEG{h}\left( \sqrt{1-\left( x+h\right) ^{2}}+% \sqrt{1-x^{2}}\right) } \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{-\left( h+2x\right) }{\sqrt{1-\left( x+h\right) ^{2}}+\sqrt{1-x^{2}}}=\dfrac{-2x}{\sqrt{1-x^{2}}+\sqrt{1-x^{2}}}=% \dfrac{-2x}{2\sqrt{1-x^{2}}}=\dfrac{-x}{\sqrt{1-x^{2}}} \end{eqnarray*} \item $f\left( x\right) =\dfrac{1}{x+2}$% \begin{eqnarray*} f^{\prime }\left( x\right) &=&\lim\limits_{h\rightarrow 0}\dfrac{f\left( x+h\right) -f\left( x\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\dfrac{1% }{x+h+2}-\dfrac{1}{x+2}}{h}=\lim\limits_{h\rightarrow 0}\dfrac{~~~\dfrac{x+2% }{\left( x+h+2\right) \left( x+2\right) }-\dfrac{x+h+2}{\left( x+h+2\right) \left( x+2\right) }~~~~}{h} \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{1}{h}\cdot \dfrac{x+2-x-h-2}{\left( x+h+2\right) \left( x+2\right) }=\lim\limits_{h\rightarrow 0}\dfrac{1}{h}% \cdot \dfrac{-h}{\left( x+h+2\right) \left( x+2\right) }=\lim\limits_{h% \rightarrow 0}\dfrac{-1}{\left( x+h+2\right) \left( x+2\right) }=-\dfrac{1}{% \left( x+2\right) ^{2}} \end{eqnarray*} \end{enumerate} \vspace{4in} {\small %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}