\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}[theorem]{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{\Large \color{blue} Lecture Notes} \chead{ \LARGE Are you ready for calculus 2? - Solutions} \rhead{\large page \thepage} \lfoot{\small \copyright $\;$ copyright Hidegkuti 2012} \rfoot{\small Last revised: January 13, 2012} \cfoot{} \textwidth 7.6in \textheight 9.6in \setlength{\headheight}{27pt} \setlength{\parindent}{0pt} \begin{document} \begin{center} {\LARGE Solutions of selected problems\bigskip } \end{center} \begin{enumerate} \item[4.] Assume that for all real numbers $x$ and $y,$% \begin{eqnarray*} \sin ^{2}x+\cos ^{2}x &=&1\text{ \ } \\ \sin \left( x+y\right) &=&\sin x\cos y+\cos x\sin y\text{ \ \ and \ } \\ \cos \left( x+y\right) &=&\cos x\cos y-\sin x\sin y \end{eqnarray*}% Prove each of the following. a) \ $\sin \left( x-y\right) =\sin x\cos y-\cos x\sin y$\newline Solution: \ We will think of $x-y$ as $x+\left( -y\right) $ and apply the sum formula to this sum.% \begin{equation*} \sin \left( x-y\right) =\sin \left( x+\left( -y\right) \right) =\sin x\cos \left( -y\right) +\cos x\sin \left( -y\right) \end{equation*}% We know that $\sin \left( -y\right) =-\sin y$ and $\cos \left( -y\right) =\cos y.$% \begin{equation*} \sin \left( x-y\right) =\sin x\cos y+\left( -1\right) \cos x\sin y=\sin x\cos y-\cos x\sin y \end{equation*}% b) \ $\cos 2x=2\cos ^{2}x-1$\newline Solution: \ \begin{equation*} \cos 2x=\cos \left( x+x\right) =\cos x\cos x-\sin x\sin x=\cos ^{2}x-\sin ^{2}x \end{equation*}% We can eliminate $\sin x$ \ using the identity $\cos ^{2}x+\sin ^{2}x=1$; by substituting $\sin ^{2}x=1-\cos ^{2}x$ \begin{equation*} \cos 2x=\cos ^{2}x-\sin ^{2}x=\cos ^{2}x-\left( 1-\cos ^{2}x\right) =2\cos ^{2}x-1 \end{equation*}% c) \ $\cos 2x=1-2\sin ^{2}x$\newline We start with $\cos 2x=\cos ^{2}x-\sin ^{2}x$ and eliminate $\cos x$ \ using the identity $\sin ^{2}x+\cos ^{2}x=1$; by substituting $\cos ^{2}x=1-\sin ^{2}x$ \begin{equation*} \cos 2x=\cos ^{2}x-\sin ^{2}x=\left( 1-\sin ^{2}x\right) -\sin ^{2}x=1-2\sin ^{2}x \end{equation*}% d) \ $\sin x=\pm \sqrt{\dfrac{1-\cos 2x}{2}}$\newline Solution: \ We start with $\cos 2x=1-2\sin ^{2}x$ and solve for $\sin x$. e) \ $\tan \left( x+y\right) =\dfrac{\tan x+\tan y}{1-\tan x\tan y}$\newline Solution: \begin{equation*} \tan \left( x+y\right) =\dfrac{\sin \left( x+y\right) }{\cos \left( x+y\right) }=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y} \end{equation*}% We will now divide both numerator and denominator by $\cos x\cos y$% \begin{eqnarray*} \tan \left( x+y\right) &=&\dfrac{\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y}}{\dfrac{\cos x\cos y-\sin x\sin y}{\cos x\cos y}}=\dfrac{\dfrac{% \sin x\cos y}{\cos x\cos y}+\dfrac{\cos x\sin y}{\cos x\cos y}}{\dfrac{\cos x\cos y}{\cos x\cos y}-\dfrac{\sin x\sin y}{\cos x\cos y}}=\dfrac{\dfrac{% \sin x}{\cos x}+\dfrac{\sin y}{\cos y}}{1-\dfrac{\sin x}{\cos x}\dfrac{\sin y% }{\cos y}} \\ &=&\dfrac{\tan x+\tan y}{1-\tan x\tan y} \end{eqnarray*}% f) \ $\sec ^{2}x=1+\tan ^{2}x$\newline Solution: \ \begin{equation*} \text{RHS}=1+\tan ^{2}x=1+\dfrac{\sin ^{2}x}{\cos ^{2}x}=\dfrac{\cos ^{2}x}{% \cos ^{2}x}+\dfrac{\sin ^{2}x}{\cos ^{2}x}=\dfrac{\cos ^{2}x+\sin ^{2}x}{% \cos ^{2}x}=\dfrac{1}{\cos ^{2}x}=\sec ^{2}x=\text{LHS} \end{equation*}% \ g) \ $\cos ^{2}x=\dfrac{1}{2}\left( \cos 2x+1\right) $\newline Solution: \ Start with $\cos 2x=2\cos ^{2}x-1$ and solve for $\cos ^{2}x$. h) \ $\dfrac{\sin x}{1+\cos x}=\dfrac{1-\cos x}{\sin x}$\newline Solution: \ \begin{equation*} \text{LHS}=\dfrac{\sin x}{1+\cos x}=\dfrac{\sin x}{1+\cos x}\cdot \dfrac{% 1-\cos x}{1-\cos x}=\dfrac{\sin x\left( 1-\cos x\right) }{1-\cos ^{2}x}=% \dfrac{\sin x\left( 1-\cos x\right) }{\sin ^{2}x}=\dfrac{1-\cos x}{\sin x}=% \text{RHS} \end{equation*} \item[5.] Simplify each of the following. a) \ $\sin \left( \sin ^{-1}x\right) $\newline Solution: \ When we compose a function with its inverse, we always get $% f\left( f^{-1}\left( x\right) \right) =x$. \ b) \ $\cos \left( \sin ^{-1}x\right) =\sqrt{1-x^{2}}$\newline Solution 1: \ We need to simplify $\cos \left( \sin ^{-1}x\right) $. \ Let $% \beta =\sin ^{-1}x$. \ We re-write the expression to be simplified:% \begin{equation*} \cos \underset{\beta }{\underbrace{\left( \sin ^{-1}x\right) }}~=~? \end{equation*}% This means that $-\dfrac{\pi }{2}<\beta <\dfrac{\pi }{2}$, $\ \sin \beta =x,$ and we need to express $\cos \beta $ in terms of $x$. \ Since $\sin ^{2}\beta +\cos ^{2}\beta =1,$% \begin{equation*} \cos \beta =\pm \sqrt{1-\sin ^{2}\beta }=\pm \sqrt{1-x^{2}} \end{equation*}% Since $-\dfrac{\pi }{2}<\beta <\dfrac{\pi }{2}$, $\cos \beta $ is positive and so $\cos \left( \sin ^{-1}x\right) =\sqrt{1-x^{2}}$ Solution 2: \ This method significantly reduces computation but can only provide us with the answer up to sign! \textbf{\ We always have to worry about the signs after the method gave us the absolute value of the answer.} We need to simplify $\cos \left( \sin ^{-1}x\right) $. \ Let $\beta =\sin ^{-1}x$. \ We re-write the expression to be simplified:% \begin{equation*} \cos \underset{\beta }{\underbrace{\left( \sin ^{-1}x\right) }}~=~? \end{equation*}% This means that $\sin \beta =x,$ and we need to express $\cos \beta $ in terms of $x$. \ Let us first draw a right triangle where $\sin \beta =x$ happens. \ One side will be labeled as $x$, another as $1,$ and one angle as $\beta $ so that it is true that $\sin \beta =x$. \ Here is such a triangle:% \FRAME{dtbpF}{2.6187in}{1.3275in}{0pt}{}{}{pic2.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.6187in;height 1.3275in;depth 0pt;original-width 3.4203in;original-height 1.7193in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'pic2.bmp';file-properties "XNPEU";}}We compute the missing side by the Pythagorean theorem and get $% \sqrt{1-x^{2}}$. \ Now we can easily compute $\cos \beta $ in terms of the triangle: $\cos \beta =\dfrac{\sqrt{1-x^{2}}}{1}=\sqrt{1-x^{2}}$. \ This is a very nice easy method, but the payback is that we need to worry about the sign of the answer: so far we only know that it is $\pm \sqrt{1-x^{2}}$. \ We know that the range of $\sin ^{-1}x$ is $\left[ -\dfrac{\pi }{2},\dfrac{% \pi }{2}\right] $. \ The cosine of all angles within this interval is non-negative, so the answer is simply $\sqrt{1-x^{2}}$. c) \ $\sin \left( \tan ^{-1}x\right) =\dfrac{x}{\sqrt{x^{2}+1}}$\newline Solution 1: \ We need to simplify $\sin \left( \tan ^{-1}x\right) $. \ Let $% \gamma =\tan ^{-1}x$. We re-write the expression to be simplified:% \begin{equation*} \sin \underset{\gamma }{\underbrace{\left( \tan ^{-1}x\right) }}~=~? \end{equation*}% \ This means that $-\dfrac{\pi }{2}<\gamma <\dfrac{\pi }{2}$, $\ \tan \gamma =x,$ and we need to express $\sin \gamma $ in terms of $x$.% \begin{eqnarray*} \sin ^{2}\gamma +\cos ^{2}\gamma &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }\sin ^{2}\gamma \\ \dfrac{\sin ^{2}\gamma }{\sin ^{2}\gamma }+\dfrac{\cos ^{2}\gamma }{\sin ^{2}\gamma } &=&\dfrac{1}{\sin ^{2}\gamma } \\ 1+\dfrac{1}{\tan ^{2}\gamma } &=&\dfrac{1}{\sin ^{2}\gamma } \\ \dfrac{\tan ^{2}\gamma +1}{\tan ^{2}\gamma } &=&\dfrac{1}{\sin ^{2}\gamma }% \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ take reciprocal} \\ \dfrac{\tan ^{2}\gamma }{\tan ^{2}\gamma +1} &=&\sin ^{2}\gamma \\ \sin \gamma &=&\pm \sqrt{\dfrac{\tan ^{2}\gamma }{1+\tan ^{2}\gamma }}=\pm \dfrac{\sqrt{\tan ^{2}\gamma }}{\sqrt{1+\tan ^{2}\gamma }}=\pm \dfrac{% \left\vert \tan \gamma \right\vert }{\sqrt{1+\tan ^{2}\gamma }}=\pm \dfrac{% \left\vert x\right\vert }{\sqrt{1+x^{2}}} \end{eqnarray*}% We can further simplify this expression by considering its sign. \ Since $-% \dfrac{\pi }{2}<\gamma <\dfrac{\pi }{2},$ the value of $\cos \gamma $ is always non-negative. \ \ This means that $\sin \gamma $ and $\tan \gamma $ have the same signs, if one is positive, so is the other; if one is negative, so is the other. \ This means that our expression can be simplified as% \begin{equation*} \sin \gamma =\dfrac{x}{\sqrt{1+x^{2}}} \end{equation*}% Solution 2: \ This method significantly reduces computation but can only provide us with the answer up to sign! \textbf{\ We always have to worry about the signs after the method gave us the absolute value of the answer.} We need to simplify $\sin \left( \tan ^{-1}x\right) $. \ Let $\gamma =\tan ^{-1}x$. \ We re-write the expression to be simplified:% \begin{equation*} \sin \underset{\gamma }{\underbrace{\left( \tan ^{-1}x\right) }}~=~? \end{equation*}% This means that $\tan \gamma =x,$ and we need to express $\sin \gamma $ in terms of $x$. \ Let us first draw a right triangle where $\tan \gamma =x$ happens. \ One side will be labeled as $x$, another as $1,$ and one angle as $\gamma $ so that it is true that $\tan \gamma =x$. \ Here is such a triangle:\FRAME{dtbpF}{2.2303in}{1.2211in}{0pt}{}{}{pic3.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.2303in;height 1.2211in;depth 0pt;original-width 3.4203in;original-height 1.8602in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'pic3.bmp';file-properties "XNPEU";}}We compute the missing side by the Pythagorean theorem and get $% \sqrt{x^{2}+1}$. \ Now we can easily compute $\sin \gamma $ in terms of the triangle: $\sin \gamma =\dfrac{x}{\sqrt{x^{2}+1}}$. \ This is a very nice easy method, but the payback is that we need to worry about the sign of the answer: so far we only know that it is $\pm \dfrac{x}{\sqrt{x^{2}+1}}$. \ We know that the range of $\tan ^{-1}x$ is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right] $. \ The cosine of all angles within this interval is non-negative, so the sign of tangent depends on the sign of sine. \ If the sine is postive, so is the tangent. \ If the sine is negative, so is the tangent. \ In the expression, $\dfrac{x}{\sqrt{x^{2}+1}},$ the denominator is clearly positive. \ The fraction is positive if $x$ is, and negative if $% x $ is. \ Since $x=\tan \gamma $ and $\sin \gamma $ must have the same sign, our expression can be simplified as $\dfrac{x}{\sqrt{x^{2}+1}}$. d) \ $\tan \left( \cos ^{-1}x\right) =\dfrac{\sqrt{1-x^{2}}}{x}$ \ \ \newline Solution 1: \ We need to simplify $\tan \left( \cos ^{-1}x\right) $. \ Let $% \theta =\cos ^{-1}x$. \ This means that $0<\theta <\pi $, $\ \cos \theta =x,$ and we need to express $\tan \theta $ in terms of $x$. We can quickly figure out $\sin \theta $ (see part b): $\sin \theta =\pm \sqrt{1-\cos ^{2}\theta }=\pm \sqrt{1-x^{2}}$. \ because $0<\theta <\pi ,$ the value of $\sin \theta $ is positive and so $\sin \theta =\sqrt{1-x^{2}}$ \ Then $\tan \theta =\dfrac{\sin \theta }{\cos \theta }=\dfrac{\sqrt{1-x^{2}}% }{x}$ Solution 2:% \begin{eqnarray*} \sin ^{2}\theta +\cos ^{2}\theta &=&1\text{ \ \ \ \ \ \ \ divide by }\cos ^{2}\theta \\ \dfrac{\sin ^{2}\theta }{\cos ^{2}\theta }+\dfrac{\cos ^{2}\theta }{\cos ^{2}\theta } &=&\dfrac{1}{\cos ^{2}\theta } \\ \tan ^{2}\theta +1 &=&\dfrac{1}{\cos ^{2}\theta } \\ \tan \theta &=&\pm \sqrt{\dfrac{1}{\cos ^{2}\theta }-1}=\pm \sqrt{\dfrac{1}{% x^{2}}-1}=\pm \sqrt{\dfrac{1-x^{2}}{x^{2}}}=\pm \dfrac{\sqrt{1-x^{2}}}{\sqrt{% x^{2}}} \\ \tan \theta &=&\pm \dfrac{\sqrt{1-x^{2}}}{\left\vert x\right\vert } \end{eqnarray*}% Since $0<\theta <\pi $, $\ $the sign of $\tan \theta $ will depend on the sign of $\cos \theta $ (clearly, $\sin \theta $ is positive on $\left( 0,\pi \right) $. \ Thus $\tan \theta $ is positive when $\cos \theta =x$ is positive and negative when $\cos \theta =x$ is negative. \ This means that our expression can be simplified as $\tan \theta =\dfrac{\sqrt{1-x^{2}}}{x}$ Solution 3: \ This method significantly reduces computation but can only provide us with the answer up to sign! \textbf{\ We always have to worry about the signs after the method gave us the absolute value of the answer.} We need to simplify $\tan \left( \cos ^{-1}x\right) $. \ Let $\delta =\tan ^{-1}x$. \ We re-write the expression to be simplified:% \begin{equation*} \tan \underset{\delta }{\underbrace{\left( \cos ^{-1}x\right) }}~=~? \end{equation*}% This means that $\cos \delta =x$, and we need to express $\tan \delta $ in terms of $x$. \ Let us first draw a right triangle where $\cos \delta =x$ happens. \ One side will be labeled as $x$, another as $1,$ and one angle as $\delta $ so that it is true that $\cos \delta =x$. \ Here is such a triangle:\FRAME{dtbpF}{2.3134in}{1.2661in}{0pt}{}{}{pic4.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.3134in;height 1.2661in;depth 0pt;original-width 3.4203in;original-height 1.8602in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'pic4.bmp';file-properties "XNPEU";}}We compute the missing side by the Pythagorean theorem and get $% \sqrt{1-x^{2}}$. \ Now we can easily compute $\tan \delta $ in terms of the triangle: $\tan \delta =\dfrac{\sqrt{1-x^{2}}}{x}$. \ This is a very nice easy method, but the payback is that we need to worry about the sign of the answer: so far we only know that it is $\pm \dfrac{\sqrt{1-x^{2}}}{x}$. \ Since $0<\theta <\pi $, $\ $the sign of $\tan \theta $ will depend on the sign of $\cos \theta $ (clearly, $\sin \theta $ is positive on $\left( 0,\pi \right) $. \ Thus $\tan \theta $ is positive when $\cos \theta =x$ is positive and negative when $\cos \theta =x$ is negative. \ This means that our expression can be simplified as $\tan \theta =\dfrac{\sqrt{1-x^{2}}}{x}$. \item[6.] Claim: $A=B=C=D=E$ where \ %TCIMACRO{\TeXButton{5 col begin}{\begin{multicols}{5}}}% %BeginExpansion \begin{multicols}{5}% %EndExpansion $A=\sqrt{\dfrac{1+\sin x}{1-\sin x}}$ $B=\dfrac{1+\sin x}{\cos x}$ $C=\sec x+\tan x$ $D=\dfrac{\cos x}{1-\sin x}$ $E=\dfrac{1}{\sec x-\tan x}$ %TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion Proof:% \begin{equation*} A=\sqrt{\dfrac{1+\sin x}{1-\sin x}}=\sqrt{\dfrac{1+\sin x}{1-\sin x}\cdot \dfrac{1+\sin x}{1+\sin x}}=\sqrt{\dfrac{\left( 1+\sin x\right) ^{2}}{1-\sin ^{2}x}}=\sqrt{\dfrac{\left( 1+\sin x\right) ^{2}}{\cos ^{2}x}}=\sqrt{\left( \dfrac{1+\sin x}{\cos x}\right) ^{2}}=\dfrac{1+\sin x}{\cos x}=B \end{equation*}% \begin{equation*} B=\dfrac{1+\sin x}{\cos x}=\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}=\sec x+\tan x=C \end{equation*}% \begin{equation*} B=\dfrac{1+\sin x}{\cos x}=\dfrac{1+\sin x}{\cos x}\cdot \dfrac{1-\sin x}{% 1-\sin x}=\dfrac{1-\sin ^{2}x}{\cos x\left( 1-\sin x\right) }=\dfrac{\cos ^{2}x}{\cos x\left( 1-\sin x\right) }=\dfrac{\cos x}{1-\sin x}=D \end{equation*}% \begin{equation*} D=\dfrac{\cos x}{1-\sin x}=\dfrac{1}{~~~\dfrac{1-\sin x}{\cos x}~~~}=\dfrac{1% }{\dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x}}=\dfrac{1}{\sec x-\tan x}=E \end{equation*} \item[7.] a) \ \ $\log _{24}90=\dfrac{\ln 90}{\ln 24}=\dfrac{\ln \left( 2\cdot 3^{2}\cdot 5\right) }{\ln \left( 2^{3}\cdot 3\right) }=\dfrac{\ln 2+\ln \left( 3^{2}\right) +\ln 5}{\ln \left( 2^{3}\right) +\ln 3}=\dfrac{\ln 2+2\ln 3+\ln 5}{3\ln 2+\ln 3}$ b) \ $2\log _{10}\left( 2x\right) +\log _{10}\left( 25x\right) -3\log _{10}0.1x=\log _{10}\dfrac{4x^{2}\left( 25x\right) }{\left( 0.1x\right) ^{3}}% =\log _{10}\left( \dfrac{100x^{3}}{\dfrac{1}{1000}x^{3}}\right) =\log _{10}10^{5}=5$ c) \ $\log _{2}3\cdot \log _{3}4\cdot \log _{4}5\cdot \log _{5}6\cdot \log _{6}7\cdot \log _{7}8=\dfrac{\ln 3}{\ln 2}\cdot \dfrac{\ln 4}{\ln 3}\cdot \dfrac{\ln 5}{\ln 4}\cdot \dfrac{\ln 6}{\ln 5}\cdot \dfrac{\ln 7}{\ln 6}% \cdot \dfrac{\ln 8}{\ln 7}=\dfrac{\ln 8}{\ln 2}=\dfrac{\ln \left( 2^{3}\right) }{\ln 2}=\dfrac{3\ln 2}{\ln 2}=3$ d) \ $\log _{3}\left\vert \tan x\right\vert =\log _{3}\left\vert \dfrac{\sin x}{\cos x}\right\vert =\log _{3}\left\vert \left( \dfrac{\cos x}{\sin x}% \right) ^{-1}\right\vert =-\log _{3}\left\vert \dfrac{\cos x}{\sin x}% \right\vert =-\log _{3}\left\vert \cot x\right\vert $ \item[10.] Claim: \ $\dfrac{d\left( x^{2}-x\right) }{dx}=2x-1$ Proof: \ \begin{eqnarray*} \dfrac{d\left( x^{2}-x\right) }{dx} &=&\lim\limits_{h\rightarrow 0}\dfrac{% \left( \left( x+h\right) ^{2}-\left( x+h\right) \right) -\left( x^{2}-x\right) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{% x^{2}+2hx+h^{2}-x-h-x^{2}+x}{h} \\ &=&\lim\limits_{h\rightarrow 0}\dfrac{2hx+h^{2}-h}{h}=\lim\limits_{h% \rightarrow 0}\dfrac{h\left( 2x+h-1\right) }{h}=\lim\limits_{h\rightarrow 0}2x+h-1=2x-1 \end{eqnarray*}% \pagebreak \item[11.] Claim: \ $f\left( x\right) =\sin ^{-1}x$ then $f^{\prime }\left( x\right) =\dfrac{1}{\sqrt{1-x^{2}}}$ Solution: \ Recall that when we compose a function $f$ with its inverse $% f^{-1,}\,$the result is always the same function:% \begin{equation*} f\left( f^{-1}\left( x\right) \right) =x \end{equation*}% We will state this fact for $f\left( x\right) =\sin x$ and differentiate both sides of the equation. \ For the left-hand side, we use the chain rule.% \begin{eqnarray*} \sin \left( \sin ^{-1}x\right) &=&x \\ \cos \left( \sin ^{-1}x\right) \cdot \dfrac{d\left( \sin ^{-1}x\right) }{dx} &=&1 \\ \dfrac{d\left( \sin ^{-1}x\right) }{dx} &=&\dfrac{1}{\cos \left( \sin ^{-1}x\right) }=\dfrac{1}{\sqrt{1-x^{2}}} \end{eqnarray*}% To prove that $\cos \left( \sin ^{-1}x\right) =\sqrt{1-x^{2}}$, see problem 5b. \item[12.] b) \ Recall that $\dint \dfrac{1}{x^{2}+1}dx=\tan ^{-1}x+C$% \begin{equation*} \dint \dfrac{x^{2}}{x^{2}+1}dx=\dint \dfrac{x^{2}+1-1}{x^{2}+1}dx=\dint \dfrac{x^{2}+1}{x^{2}+1}-\dfrac{1}{x^{2}+1}dx=\dint 1-\dfrac{1}{x^{2}+1}% dx=\dint 1dx-\dint \dfrac{1}{x^{2}+1}dx=x-\tan ^{-1}x+C \end{equation*} \end{enumerate} \bigskip \vspace{1in} \vspace{2in} \vspace{2in} {\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}