\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=0.6in,right=0.7in,top=0.7in,bottom=0.7in} \pagestyle{fancy} \lhead{\color{blue} \large Lecture Notes} \chead{\color{black} \LARGE Differentiating Trigonometric Functions} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ copyright Hidegkuti, Powell, 2012} \rfoot{\small Last revised: August 7, 2012} \textwidth 7.5in \textheight 9.6in \setlength{\headheight}{25pt} \setlength{\parindent}{0in} \begin{document} \bigskip \bigskip %TCIMACRO{\TeXButton{2col begin}{\begin{multicols}{2}}}% %BeginExpansion \begin{multicols}{2}% %EndExpansion \begin{enumerate} \item $\dfrac{d}{dx}\sin x=\cos x\medskip \smallskip $\bigskip $\medskip $ \item $\dfrac{\smallskip d}{dx}\cos x=-\sin x\medskip \bigskip \smallskip $ \item $\dfrac{d}{dx}\tan x=\sec ^{2}x=\tan ^{2}x+1\medskip \bigskip \smallskip \medskip $ \item $\dfrac{d}{dx}\cot x=-\csc ^{2}x=-\cot ^{2}x-1\bigskip \medskip \smallskip $ \item $\dfrac{d}{dx}\sec x=\dfrac{\sin x}{\cos ^{2}x}=\sec x\tan x\medskip \smallskip $\bigskip \item $\dfrac{d}{dx}\csc x=-\dfrac{\cos x}{\sin ^{2}x}=-\csc x\cot x\bigskip \medskip \smallskip $ \item $\dfrac{d}{dx}\sin ^{-1}x=\dfrac{1}{\sqrt{1-x^{2}}}\medskip $\bigskip \item $\dfrac{d}{dx}\cos ^{-1}x=-\dfrac{1}{\sqrt{1-x^{2}}}\bigskip \medskip $ \item $\dfrac{d}{dx}\tan ^{-1}x=\dfrac{1}{x^{2}+1}\medskip $\bigskip \item $\dfrac{d}{dx}\cot ^{-1}x=-\dfrac{1}{x^{2}+1}\bigskip \medskip $ \item $\dfrac{d}{dx}\sec ^{-1}x=\dfrac{1}{\left\vert x\right\vert \sqrt{% x^{2}-1}}\bigskip \medskip $ \item $\dfrac{d}{dx}\csc ^{-1}x=-\dfrac{1}{\left\vert x\right\vert \sqrt{% x^{2}-1}}\medskip $ \ \ \end{enumerate} %TCIMACRO{\TeXButton{m col end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \pagebreak \begin{center} {\LARGE Proofs}\bigskip \end{center} \fbox{Theorems 1 and 2: \ $\dfrac{d}{dx}\sin x=\cos x$ \ and $\dfrac{d}{dx}% \cos x=-\sin x$}\bigskip Claim 1.) \ $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=1\medskip $ Proof: \ This theorem and the next one are necessary for differentiating $% \sin x$ and $\cos x$. \ Recall a theorem: \ Let $\ r$ be the radius of a circle. \ If $\alpha $ is measured in radians, then the area of a sector with a central angle of $\alpha $ is $A_{\text{sector}}=\dfrac{\alpha r^{2}}{% 2}$. \ (Notation: \ $\overline{AB}$ will denote the length of line segment $% AB$.)$\medskip $ Let $x$ be a very small positive angle, measured in radians, drawn into a unit circle as shown on the picture below. \ Let $B$ be the point where the unit circle intersects the ray determined by $x$. \ We then draw a tangent line to the circle at point $B$. \ Let $A$ be the point where the tangent line intersects the $x-$axis. \ We also draw a vertical line through $B.$ \ Let $D$ be the point where this vertical line intersects the $x-$axis. \ Finally, let us denote by $E$ the point with coordinates $\left( 0,1\right) $% .\FRAME{dtbpF}{2.776in}{2.2399in}{0pt}{}{}{sinx_xbetter.bmp}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.776in;height 2.2399in;depth 0pt;original-width 2.7337in;original-height 2.2001in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'sinx_xbetter.bmp';file-properties "XNPEU";}} The proof will be based on the following fact: because they include each other, the following three areas can be easily compared: \begin{equation*} \text{Area of triangle }CDB\text{ }\leq \text{ Area of sector }CEB\text{ }% \leq \text{ Area of triangle }ABC \end{equation*} Area of triangle $CDB$: \ the horizontal side, $\overline{CD}=\cos x$ and the vertical side, $\overline{DB}=\sin x$. \ Since this is a right triangle, the area is: $A_{CDB}=\dfrac{1}{2}\sin x\cos x\medskip $ Area of sector $CEB$: $A_{\text{sector}}=\dfrac{1^{2}x}{2}=\dfrac{x}{2}% \medskip $ Area of triangle $ABC$: there is a right angle at point $B$ because the tangent line drawn to a circle is perpendicular to the radius drawn to the point of tangency. \ So the area is $A_{ABC}=\dfrac{1}{2}\overline{AB}\cdot \overline{BC}$. \ Clearly $\overline{BC}=1$. \ To compute $\overline{AB}$, in triangle $ABC$, \ $\tan x=\dfrac{\overline{AB}}{1}$ and so $\overline{AB}% =\tan x$.$\medskip $ Area of triangle $ABC$: \ $\dfrac{1}{2}\left( 1\right) \left( \tan x\right) =% \dfrac{\tan x}{2}$ or $\dfrac{\sin x}{2\cos x}$. \ So now \begin{equation*} \text{Area of triangle }CDB\text{ }\leq \text{ Area of sector }CEB\text{ }% \leq \text{ Area of triangle }ABC \end{equation*}% translates to \begin{equation*} \dfrac{1}{2}\sin x\cos x\leq \dfrac{x}{2}\leq \dfrac{\sin x}{2\cos x} \end{equation*} Let us divide all three sides by $\dfrac{\sin x}{2}$. \ Because $x$ is small and positive, $\dfrac{\sin x}{2}$ is positive and so we do not need to reverse the inequality signs.% \begin{equation*} \cos x\leq \dfrac{x}{\sin x}\leq \dfrac{1}{\cos x} \end{equation*}% Suppose now that $x$ approaches zero. \ Then both $\cos x$ and $\dfrac{1}{% \cos x}$ approach $1$. \ By the sandwich principle, $\dfrac{x}{\sin x}$, the quantity locked in between those two must also approach $1.$ \ \begin{equation*} \begin{array}{ccccc} \cos x & \leq & \dfrac{x}{\sin x} & \leq & \dfrac{1}{\cos x} \\ \downarrow & & & & \downarrow \\ 1 & & & & 1% \end{array}% \end{equation*}% If $\dfrac{x}{\sin x}$ approaches $1,$ so is its reciprocal, $\dfrac{\sin x}{% x}$.$\medskip $ So far, we have proven the statement for positive values of $x$, that is, $% \lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin x}{x}=1$. \ A similar argument works for negative values of $x$. \bigskip $\medskip \medskip $ Claim 2.) \ $\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x}=0\medskip $ Proof: \begin{eqnarray*} \lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x} &=&\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1}{x}\cdot 1=\lim\limits_{x\rightarrow 0}\left( \dfrac{\cos x-1}{x}\cdot \dfrac{\cos x+1}{\cos x+1}\right) =\lim\limits_{x\rightarrow 0}% \dfrac{\cos ^{2}x-1}{x\left( \cos x+1\right) }=\lim\limits_{x\rightarrow 0}% \dfrac{-\left( 1-\cos ^{2}x\right) }{x\left( \cos x+1\right) } \\ &=&\lim\limits_{x\rightarrow 0}\dfrac{-\sin ^{2}x}{x\left( \cos x+1\right) }% =\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}\cdot \dfrac{-\sin x}{\cos x+1}% =\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}\cdot \lim\limits_{x\rightarrow 0}\dfrac{-\sin x}{\cos x+1}=1\cdot 0=0 \end{eqnarray*} We are now ready to prove that \ $\dfrac{d}{dx}\sin x=\cos x$ \ and \ $% \dfrac{d}{dx}\cos x=-\sin x\medskip $ Proof:% \begin{eqnarray*} \dfrac{d}{dx}\sin x &=&\lim\limits_{h\rightarrow 0}\dfrac{\sin \left( x+h\right) -\sin x}{h}=\lim\limits_{h\rightarrow 0}\dfrac{\sin x\cos h+\cos x\sin h-\sin x}{h} \\ &=&\lim\limits_{h\rightarrow 0}\left( \dfrac{\sin x\cos h-\sin x}{h}+\dfrac{% \cos x\sin h}{h}\right) \medskip \medskip =\lim\limits_{h\rightarrow 0}% \dfrac{\sin x\left( \cos h-1\right) }{h}+\lim\limits_{h\rightarrow 0}\cos x% \dfrac{\sin h}{h} \\ &=&\sin x\lim\limits_{h\rightarrow 0}\dfrac{\cos h-1}{h}+\cos x\lim\limits_{h\rightarrow 0}\dfrac{\sin h}{h}=\sin x\cdot 0+\cos x\cdot 1=\cos x \end{eqnarray*}% \begin{eqnarray*} \dfrac{d}{dx}\cos x &=&\lim\limits_{h\rightarrow 0}\dfrac{\cos \left( x+h\right) -\cos x}{h}=\lim\limits_{h\rightarrow 0}\dfrac{\cos x\cos h-\sin x\sin h-\cos x}{h} \\ &=&\lim\limits_{h\rightarrow 0}\left( \dfrac{\cos x\cos h-\cos x}{h}-\dfrac{% \sin x\sin h}{h}\right) =\lim\limits_{h\rightarrow 0}\dfrac{\cos x\left( \cos h-1\right) }{h}-\lim\limits_{h\rightarrow 0}\dfrac{\sin x\sin h}{h} \\ &=&\cos x\lim\limits_{h\rightarrow 0}\dfrac{\cos h-1}{h}-\sin x\lim\limits_{h\rightarrow 0}\dfrac{\sin h}{h}=\cos x\cdot 0-\sin x\cdot 1=-\sin x \end{eqnarray*}% \bigskip \pagebreak \fbox{Theorem 3 and 4: \ $\dfrac{d}{dx}\tan x=\sec ^{2}x=\tan ^{2}x+1$ and $% \dfrac{d}{dx}\cot x=-\csc ^{2}x=-\cot ^{2}x-1$.}\bigskip Proof: \ We write $\tan x=\dfrac{\sin x}{\cos x}$ and apply the quotient rule.% \begin{equation*} \dfrac{d}{dx}\left( \dfrac{\sin x}{\cos x}\right) =\dfrac{\left( \dfrac{d}{dx% }\sin x\right) \cos x-\left( \dfrac{d}{dx}\cos x\right) \sin x}{\cos ^{2}x}=% \dfrac{\cos x\cos x-\left( -\sin x\right) \sin x}{\cos ^{2}x}=\dfrac{\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}=\dfrac{1}{\cos ^{2}x}=\sec ^{2}x \end{equation*}% We will now prove $\dfrac{1}{\cos ^{2}x}=\tan ^{2}x+1,$ which is a very important connection. \ Looking at the previous computation, \begin{equation*} \dfrac{1}{\cos ^{2}x}=\dfrac{\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}=\dfrac{\cos ^{2}x}{\cos ^{2}x}+\dfrac{\sin ^{2}x}{\cos ^{2}x}=1+\tan ^{2}x \end{equation*} The proof for$\dfrac{d}{dx}\cot x=-\cot ^{2}x-1=-\csc ^{2}x$ is very similar. \ We apply the quotient rule.% \begin{eqnarray*} \dfrac{d}{dx}\cot x &=&\dfrac{d}{dx}\left( \dfrac{\cos x}{\sin x}\right) =% \dfrac{\left( \dfrac{d}{dx}\cos x\right) \sin x-\cos x\left( \dfrac{d}{dx}% \sin x\right) }{\sin ^{2}x}=\dfrac{-\sin x\sin x-\cos x\left( \cos x\right) }{\sin ^{2}x}=\dfrac{-\sin ^{2}x-\cos ^{2}x}{\sin ^{2}x} \\ &=&-\dfrac{1}{\sin ^{2}x}=-\csc ^{2}x \end{eqnarray*}% Also, \begin{equation*} \dfrac{-\sin ^{2}x-\cos ^{2}x}{\sin ^{2}x}=\dfrac{-\sin ^{2}x}{\sin ^{2}x}-% \dfrac{\cos ^{2}x}{\sin ^{2}x}=-1-\cot ^{2}x \end{equation*}% \bigskip \fbox{Theorems \ 5 and 6: $\dfrac{d}{dx}\sec x=\sec x\tan x$ \ and \ $\dfrac{% d}{dx}\csc x=-\csc x\cot x$}\bigskip Proof: \ We write $\sec x=\dfrac{1}{\cos x}=\left( \cos x\right) ^{-1}$ and apply the chain rule.% \begin{equation*} \dfrac{d}{dx}\sec x=\dfrac{d}{dx}\left( \cos x\right) ^{-1}=-1\left( \cos x\right) ^{-2}\left( \dfrac{d}{dx}\cos x\right) =\dfrac{-1}{\cos ^{2}x}% \left( -\sin x\right) =\dfrac{\sin x}{\cos ^{2}x}=\dfrac{1}{\cos x}\cdot \dfrac{\sin x}{\cos x}=\sec x\tan x \end{equation*} The proof for\ $\dfrac{d}{dx}\csc x$ is virtually identical: \ we apply the chain rule.% \begin{equation*} \dfrac{d}{dx}\csc x=\dfrac{d}{dx}\left( \sin x\right) ^{-1}=-1\left( \sin x\right) ^{-2}\left( \dfrac{d}{dx}\sin x\right) =\dfrac{-1}{\sin ^{2}x}\cos x=-\dfrac{\cos x}{\sin ^{2}x}=-\dfrac{\cos x}{\sin x}\dfrac{1}{\sin x}=-\cot x\csc x \end{equation*}% \bigskip Note: why do we prefer the form $\sec x\tan x$ over the form $\dfrac{\sin x}{% \cos ^{2}x}$? \ One of the reasons is the adventage we'll see in differentiating the inverse functions $\sec ^{-1}x$ and $\csc ^{-1}x$% .\bigskip \pagebreak \fbox{Theorems 7 and 8: \ \ $\dfrac{d}{dx}\sin ^{-1}x=\dfrac{1}{\sqrt{1-x^{2}% }}$ and $\dfrac{d}{dx}\cos ^{-1}x=-\dfrac{1}{\sqrt{1-x^{2}}}$}$\medskip $ Proof: \ $\medskip $ Claim 1: $\cos \left( \sin ^{-1}x\right) =\sqrt{1-x^{2}}\medskip $\newline We first introduce a new variable, $\beta $. \ Let $\beta =\sin ^{-1}x$. \ This means that $-\dfrac{\pi }{2}<\beta <\dfrac{\pi }{2}$, and $\sin \beta =x $. \ We need to simplify $\cos \underset{\beta }{\underbrace{\left( \sin ^{-1}x\right) }}=\cos \beta $. \ 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 $\cos \left( \sin ^{-1}x\right) =\sqrt{1-x^{2}}$. \ $\medskip $ 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}{dx}\sin ^{-1}x &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }\cos \left( \sin ^{-1}x\right) \\ \dfrac{d}{dx}\sin ^{-1}x &=&\dfrac{1}{\cos \left( \sin ^{-1}x\right) }=% \dfrac{1}{\sqrt{1-x^{2}}} \end{eqnarray*}% \bigskip The proof for \ $\dfrac{d}{dx}\cos ^{-1}x=-\dfrac{1}{\sqrt{1-x^{2}}}$ is virtually identical.$\medskip $ Proof: \ Claim 1: $\ \sin \left( \cos ^{-1}x\right) =\sqrt{1-x^{2}}$. Let $\alpha =\cos ^{-1}x$. \ Then $x=\cos \alpha $ and $\alpha $ is between $% 0$ and $\pi $. \ Then \begin{equation*} \sin \left( \underset{\alpha }{\underbrace{\cos ^{-1}x}}\right) =\sin \alpha =\pm \sqrt{1-\cos ^{2}\alpha }=\pm \sqrt{1-x^{2}} \end{equation*}% Since $\alpha $ is between $0$ and $\pi $, $\sin \alpha $ is positive and so $\sin \alpha =\sqrt{1-x^{2}}$.$\medskip $ 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 $f\left( x\right) =\cos x$ and differentiate both sides of the equation. \ For the left-hand side, we use the chain rule.% \begin{eqnarray*} \cos \left( \cos ^{-1}x\right) &=&x \\ -\sin \left( \cos ^{-1}x\right) \cdot \dfrac{d}{dx}\cos ^{-1}x &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }\sin \left( \cos ^{-1}x\right) \\ \dfrac{d}{dx}\cos ^{-1}x &=&-\dfrac{1}{\sin \left( \cos ^{-1}x\right) }=-% \dfrac{1}{\sqrt{1-x^{2}}} \end{eqnarray*}% $\bigskip $ \fbox{Theorems 9 and 10: $\ \dfrac{d}{dx}\tan ^{-1}x=\dfrac{1}{x^{2}+1}$ and $\dfrac{d}{dx}\cot ^{-1}x=-\dfrac{1}{x^{2}+1}$}\bigskip Proof: \ Recall that $\dfrac{d}{dx}\tan x=\sec ^{2}x=\tan ^{2}x+1$. \ Also 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 $f\left( x\right) =\tan x$ and differentiate both sides of the equation. \ For the left-hand side, we use the chain rule.% \begin{eqnarray*} \tan \left( \tan ^{-1}x\right) &=&x \\ \sec ^{2}\left( \tan ^{-1}x\right) \cdot \dfrac{d}{dx}\tan ^{-1}x &=&1 \\ \left( \tan ^{2}\left( \tan ^{-1}x\right) +1\right) \cdot \dfrac{d}{dx}\tan ^{-1}x &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\tan \left( \tan ^{-1}x\right) =x \\ \left( x^{2}+1\right) \cdot \dfrac{d}{dx}\tan ^{-1}x &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }x^{2}+1 \\ \dfrac{d}{dx}\tan ^{-1}x &=&\dfrac{1}{x^{2}+1} \end{eqnarray*}% \bigskip The proof for $\dfrac{d}{dx}\cot ^{-1}x=-\dfrac{1}{x^{2}+1}$ is virtually identical. \ We compose the function $\cot x$ with its inverse $\cot ^{-1}x\, $ and differentiate. \ Recall that $\dfrac{d}{dx}\cot x=-\cot ^{2}x-1$% \begin{eqnarray*} \cot \left( \cot ^{-1}x\right) &=&x \\ \left( -\cot ^{2}\left( \cot ^{-1}x\right) -1\right) \cdot \dfrac{d}{dx}\cot ^{-1}x &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\cot \left( \cot ^{-1}x\right) =x \\ \left( -x^{2}-1\right) \cdot \dfrac{d}{dx}\cot ^{-1}x &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }-x^{2}-1 \\ \dfrac{d}{dx}\cot ^{-1}x &=&\dfrac{1}{-x^{2}-1}=-\dfrac{1}{x^{2}+1} \end{eqnarray*}% $\bigskip \bigskip $ \fbox{Theorem 11 and 12: $\ \dfrac{d}{dx}\sec ^{-1}x=\dfrac{1}{\left\vert x\right\vert \sqrt{x^{2}-1}}$ \ and \ $\dfrac{d}{dx}\csc ^{-1}x=-\dfrac{1}{% \left\vert x\right\vert \sqrt{x^{2}-1}}$}$\bigskip $ Proof: \ We compose the function $\sec x$ with its inverse $\sec ^{-1}x$ and differentiate. \ Recall that $\dfrac{d}{dx}\sec x=\sec x\tan x$.% \begin{eqnarray*} \sec \left( \sec ^{-1}x\right) &=&x \\ \sec \left( \sec ^{-1}x\right) \tan \left( \sec ^{-1}x\right) \cdot \dfrac{d% }{dx}\sec ^{-1}x &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\sec \left( \sec ^{-1}x\right) =x \\ x\tan \left( \sec ^{-1}x\right) \cdot \dfrac{d}{dx}\sec ^{-1}x &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }x\tan \left( \sec ^{-1}x\right) \\ \dfrac{d}{dx}\sec ^{-1}x &=&\dfrac{1}{x\tan \left( \sec ^{-1}x\right) } \end{eqnarray*} To simplify $\tan \left( \sec ^{-1}x\right) $, we introduce a new variable $% \alpha $. \ Let $\alpha =\sec ^{-1}x$. \ The we have $\tan \underset{\alpha }% {\left( \underbrace{\sec ^{-1}x}\right) }=\tan \alpha $ where $\sec \alpha =x $ and $\alpha $ is between $0$ and $\pi $. \ Recall that $\sec ^{2}\alpha =\tan ^{2}\alpha +1$. \ If we don't have this formula memorized, we can easily derive it from the Pythagorean identity:% \begin{eqnarray*} \sin ^{2}\alpha +\cos ^{2}\alpha &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }\cos ^{2}x \\ \tan ^{2}\alpha +1 &=&\sec ^{2}\alpha \\ \tan \alpha &=&\pm \sqrt{\sec ^{2}\alpha -1} \end{eqnarray*}% Thus the derivative is \begin{equation*} \dfrac{d}{dx}\sec ^{-1}x=\dfrac{1}{x\left( \pm \sqrt{x^{2}-1}\right) }=\pm \dfrac{1}{x\sqrt{x^{2}-1}} \end{equation*}% We now need to figure out the sign of the derivative. \ From the graph of $% \sec ^{-1}x$ we can see that it is strictly increasing on both intervals making up its domain, thus the derivative is always positive. \ If $x$ is positive, then $\dfrac{d}{dx}\sec ^{-1}x=\dfrac{1}{x\sqrt{x^{2}-1}}$ and if $% x$ is negative, then $\dfrac{d}{dx}\sec ^{-1}x=-\dfrac{1}{x\sqrt{x^{2}-1}}$. \ \ This can be expressed in a shorter form as \begin{equation*} \dfrac{d}{dx}\sec ^{-1}x=\dfrac{1}{\left\vert x\right\vert \sqrt{x^{2}-1}} \end{equation*}% \bigskip The proof for $\dfrac{d}{dx}\csc ^{-1}x=-\dfrac{1}{\left\vert x\right\vert \sqrt{x^{2}-1}}$ is virtually identical. \ \ As before, we compose the function $\csc x$ with its inverse and differentiate. \ Recall that $\dfrac{d% }{dx}\csc x=-\csc x\cot x$% \begin{eqnarray*} \csc \left( \csc ^{-1}x\right) &=&x \\ -\csc \left( \csc ^{-1}x\right) \cot \left( \csc ^{-1}x\right) \cdot \dfrac{d% }{dx}\csc ^{-1}x &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\csc \left( \csc ^{-1}x\right) =x \\ -x\cot \left( \csc ^{-1}x\right) \cdot \dfrac{d}{dx}\csc ^{-1}x &=&1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }-x\cot \left( \csc ^{-1}x\right) \\ \dfrac{d}{dx}\csc ^{-1}x &=&-\dfrac{1}{x\cot \left( \csc ^{-1}x\right) } \end{eqnarray*} Simplify $\cot \underset{\alpha }{\left( \underbrace{\csc ^{-1}x}\right) }% =\cot \alpha $ where $\csc \alpha =x$ and $\alpha $ is between $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$.% \begin{eqnarray*} \sin ^{2}\alpha +\cos ^{2}\alpha &=&1\text{ \ \ \ \ \ \ \ \ divide by }\sin ^{2}x \\ 1+\cot ^{2}\alpha &=&\csc ^{2}\alpha \\ \cot ^{2}\alpha &=&\csc ^{2}\alpha -1 \\ \cot \alpha &=&\pm \sqrt{\csc ^{2}\alpha -1} \end{eqnarray*}% Thus the derivative is \begin{equation*} \dfrac{d}{dx}\csc ^{-1}x=-\dfrac{1}{x\left( \pm \sqrt{x^{2}-1}\right) }=\pm \dfrac{1}{x\sqrt{x^{2}-1}} \end{equation*}% From the graph of $\csc ^{-1}x$ we can see that it is strictly decreasing on both intervals of its domain, thus the derivative is always negative. \ If $x $ is positive, then $\dfrac{d}{dx}\csc ^{-1}x=-\dfrac{1}{x\sqrt{x^{2}-1}}$ and if $x$ is negative, then $\dfrac{d}{dx}\csc ^{-1}x=\dfrac{1}{x\sqrt{% x^{2}-1}}$. \ \ This can be expressed in a shorter form as \begin{equation*} \dfrac{d}{dx}\csc ^{-1}x=-\dfrac{1}{\left\vert x\right\vert \sqrt{x^{2}-1}} \end{equation*}% $\medskip $ \bigskip \href{http://www.teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html% }{For more documents like this, visit our page at\ http://www.teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}