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\institution{SIAST Palliser Campus}
\department{Mathematics}
\course{MAT 226}
\unittitle{Derivatives of Transcendental Functions}
\unitnumber{5}
\date{January 2011}
\title{STUDY GUIDE}
\author{Robert G. Petry}
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\publisher{the Department of Mathematics, SIAST Palliser Campus}
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\section{Derivatives of Trigonometric Functions}
In this course we have sufficient rules to differentiate any function we
can imagine except for special functions like $\sin x$ or $e^x$.  In order
to differentiate such functions one has to appeal to the fundamental definition
of the derivative:
\[\frac{dy}{dx}=\lim_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}
=\lim_{\Delta x\rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}\ ,\]
and carefully consider the properties of the function in question.
For instance, one may use geometric considerations (see Text p. 943)
to determine that the derivative of $y=\sin x$ is:
\[\frac{dy}{dx}=\frac{d(\sin x)}{dx}=\cos x\]
Here and in the other formulas, the angular variable $x$
\textbf{must be in radians}.  The derivative of $y=\cos x$ is:
\[\frac{dy}{dx}=\frac{d(\cos x)}{dx}=-\sin x\]
Note the minus sign in the last equation.  Using these two relations,
trig identities, and our basic rules of differentiation one can find
other trigonometric derivatives as shown in the following example.

\textbf{Example}

Find the derivative of $y=\tan x$. (Hint: Use that $\tan x=\frac{\sin
  x}{\cos x}$\ .)
\clearpage
The following table summarizes the trig function derivatives.

\begin{boxedeqnarray2}
\frac{d}{dx}\left(\sin x\right) & = & \cos x \\
\frac{d}{dx}\left(\cos x\right) & = & -\sin x \\
\frac{d}{dx}\left(\tan x\right) & = & \sec^2 x \\
\frac{d}{dx}\left(\cot x\right) & = & -\csc^2 x \\
\frac{d}{dx}\left(\sec x\right) & = & \sec x\tan x \\
\frac{d}{dx}\left(\csc x\right) & = & -\csc x\cot x
\end{boxedeqnarray2}

Often these formulas will be written more generally as (for instance):
\[\frac{d}{dx}\left(\sin u\right) = \cos u\cdot \frac{du}{dx}\]
to indicate that the chain rule must be applied if the argument of the
trig function differs from the variable which is being differentiated.
Such notation is shown in the chain rule versions of our derivative
formulas on the formula sheet.  One must remember that the chain rule
($\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$) is required if the
argument does differ from the variable differentiated.  For instance,
if $y=\sin(x^2)$ then
\[\frac{dy}{dx}=\frac{d}{dx}\underbrace{\left(\sin x^2\right)}_{y=\sin u}=\underbrace{\left(\cos x^2\right)}_{\frac{dy}{du}=\cos u}\cdot\underbrace{\frac{d}{dx}(x^2)}_{\frac{du}{dx}}=\left(\cos x^2\right)\cdot(2x)=2x\cos x^2\]
Note that here $\sin x^2 = \sin(x^2)$ not $(\sin x)^2$\ . The latter
we would write $\sin^2 x$.

These trigonometric derivatives are still just derivatives.  All the
rules of derivatives, higher order derivatives, implicit
differentiation, tangents, extreme values, rates of change, related
rates, and max-min problems can be applied to such functions. Examples
of each of these can be found in the assignment and the problems at
the end of the unit.

\textassignment{Sec. 33-2}{Ex. 2 (P. 951) \# 1-41 (odd)}

\section{Derivatives of Inverse Trigonometric Functions}
As with trigonometric functions, it is desirable to find derivatives of
\textbf{inverse} trigonometric functions like $\arcsin x $ (also
written $\sin^{-1}x$).  Since if $y=\arcsin x$ one has that
\[\sin y=\sin(\arcsin x)=x\]
one can use implicit differentiation of the latter to find the
derivative of $\arcsin$.

\textbf{Example}

Find the derivative $\frac{d}{dx}(\arcsin x)$\ . (Hint: Also use that
if $\sin y = x = \frac{x}{1}$, then by the Pythagorean theorem $\cos
y=\sqrt{1-x^2}$.)
\clearpage
The following table summarizes the inverse trig function derivatives.

\begin{boxedeqnarray2}
\frac{d}{dx}\left(\sin^{-1} x\right) & = & \frac{1}{\sqrt{1-x^2}} \\
\frac{d}{dx}\left(\cos^{-1} x\right) & = & \frac{-1}{\sqrt{1-x^2}} \\
\frac{d}{dx}\left(\tan^{-1} x\right) & = & \frac{1}{1+x^2} \\
\frac{d}{dx}\left(\cot^{-1} x\right) & = & \frac{-1}{1+x^2} \\
\frac{d}{dx}\left(\sec^{-1} x\right) & = & \frac{1}{|x|\sqrt{x^2-1}} \\
\frac{d}{dx}\left(\csc^{-1} x\right) & = & \frac{-1}{|x|\sqrt{x^2-1}}
\end{boxedeqnarray2}

Note that for $\sin^{-1}$ and $\cos^{-1}$ the derivative is only
meaningfully defined where the function is, namely $|x|<1$.  For
$\sec^{-1}$ and $\csc^{-1}$ one requires $|x|>1$. Finally $\tan^{-1}$
and $\cot^{-1}$ have derivatives defined for all $x$.

As with the trigonometric derivatives, one simply uses these rules
to differentiate any function that involves an inverse trig function.
Examples may be found in the assignment and at the end of the unit.

\textassignment{Sec. 33-3}{Ex. 3 (P. 956) \# 1-21 (odd)}

\section{Derivatives of Logarithmic Functions}

The derivative of the natural logarithm $\ln x$ gives the following
surprisingly simple result: 
\begin{boxedequation}
\frac{d}{dx}\left(\ln x\right)=\frac{1}{x}
\end{boxedequation}
(See Text p. 956 for proof.)

Logarithms to bases other than $e$ (the base of $\ln=\log_e$),
introduce a constant factor of $\ln b$ in the denominator:
\begin{boxedequation}
\frac{d}{dx}\left(\log_b x\right)=\frac{1}{x\ln b}
\end{boxedequation}

These rules can be used to differentiate logarithmic functions as they
arise in any differentiation problem as shown in the assignment below
and at the end of the unit.

\textassignment{Sec. 33-4}{Ex. 4 (P. 960) \# 1-51 (odd)}

\section{Derivatives of Exponential Functions}

The derivative of the exponential function of base $e$ is very special
indeed as it is the only functions whose derivative is itself!
\begin{boxedequation}
\frac{d}{dx}\left(e^x\right)=e^x
\end{boxedequation}

\textbf{Example}

Prove that the derivative of $e^x$ is itself. (Hint: Use that if
$y=e^x$ then $x=\ln y$ and implicitly differentiate.)
\vspace{3in}

For exponentials involving bases other than $e$ one has to introduce
the constant $\ln b$ as with logarithms:
\begin{boxedequation}
\frac{d}{dx}\left(b^x\right)=b^x\ln b
\end{boxedequation}

Examples using these derivatives may be found in the assignment and at
the end of the unit.

\textassignment{Sec. 33-5}{Ex. 5 (P. 964) \# 1-61 (odd)}

\section{General Questions}
Use calculus techniques and the new derivative formulas from
this unit to solve the following problems.
\begin{enumerate}
\item Find the derivatives of the following:
\begin{enumerate}
\item $\displaystyle y=x^2+\sin x+\cos^{-1}x-\ln x+5^x$
\item $\displaystyle y=\sin x\cos x$
\item $\displaystyle y=\frac{\ln x}{x}$
\item $\displaystyle y=\tan^2 x$
\item $\displaystyle y=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$
\item $\displaystyle y=2\sqrt[3]{\sec^{-1}x}$
\end{enumerate}
\item If $f(x)=\sin^{-1}(x)$, find $f''(x)$ and $f''(0)$.
\item Find the derivative $\frac{dy}{dx}$ of the implicitly defined function:
\[\ln y+x^3=x^2\cos y \]
\item Find the minima, maxima and inflection points of the function $y=\sin^2\theta$ for $0\le\theta<2\pi$\ .
\item The elevation above sea level of a hill (in km) is described by the
  function \mbox{$y=xe^{-3x}$}.  Find the horizontal distance $x$ at which the
  hill is a maximum as well the maximum elevation of the hill.
\item Find the horizontal distance $x$ at which the hill is maximum
  from the last problem using \emph{Newton's method}.  (Hint: Since
  you will be solving $y'=0$, you will want to call $f(x)$ in Newton's
  method your expression for $y'$.  Then $f'(x)$ in the method will be
  your expression for $y''$.)
\item An airplane flies horizontally away from you at a constant
  altitude of 10.0 km.  When the plane is at a ground distance
  (horizontal) from you of 5.00 km find the rate of change of the
  angle of inclination of the plane from your point of observation if
  the plane is travelling at a speed of $25\bar{0}$ m/s at that
  moment.
\end{enumerate}
\clearpage
\section*{Solutions}
\begin{enumerate}
\item \begin{enumerate}
\item $\displaystyle \frac{dy}{dx}=2x+\cos x-\frac{1}{\sqrt{1-x^2}}-\frac{1}{x}+5^x\ln 5$
\item $\displaystyle \frac{dy}{dx}=-\sin^2x+\cos^2x$
\item $\displaystyle \frac{dy}{dx}=\frac{1-\ln x}{x^2}$
\item $\displaystyle \frac{dy}{dx}=2\tan x\sec^2 x$
\item $\displaystyle \frac{dy}{dx}=-\frac{x}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$
\item $\displaystyle \frac{dy}{dx}=\frac{2}{3}\left(\sec^{-1}x\right)^{-\frac{2}{3}}\frac{1}{|x|\sqrt{x^2-1}}$
\end{enumerate}
\item $\displaystyle f''(x)=x(1-x^2)^{-\frac{3}{2}}$, $\displaystyle f''(0)=0$
\item $\displaystyle y'=\frac{2x\cos y-3x^2}{\frac{1}{y}+x^2\sin y}$
\item \begin{description}
\item[relative minima:] $0$, $\pi$
\item[relative maxima:] $\displaystyle \frac{\pi}{2}$, $\displaystyle
  \frac{3\pi}{2}$
\item[inflection points:] $\displaystyle \frac{\pi}{4}$, $\displaystyle \frac{3\pi}{4}$, $\displaystyle \frac{5\pi}{4}$, $\displaystyle \frac{7\pi}{4}$
\end{description}
\item \begin{description}
\item[horizontal distance:] $\displaystyle \frac{1}{3}\ \mbox{km}$
\item[maximum height:] $\displaystyle \frac{1}{3}e^{-1}=.123\ \mbox{km}$.
\end{description}
\item $\displaystyle x=.333333333=\frac{1}{3}\ \mbox{km}$
\item $\displaystyle \left.\frac{d\theta}{dt}\right|_{x=5\ \mbox{\tiny km}}=-.0200\frac{\mbox{\small rad}}{\mbox{\small s}}$
\end{enumerate}
\clearpage
\section{Further Review Problems}
Chapter 33, page 966,  \# 1-25 (7 \& 16 require implicit differentiation.)  The following
answers are for the even numbered problems:
\begin{description}
\item[2.] $\displaystyle 2\ln{5}\left(5^{2x+3}\right)$
\item[4.] $\displaystyle 2\sec^2{x}\tan{x}$
\item[6.] $\displaystyle -\left(\arcsin{2x}\right)^{-\frac{3}{2}}\left(1-4x^2\right)^{-\frac{1}{2}}$
\item[8.] $\displaystyle e^{2x}\left(2x+1\right)$
\item[10.] $\displaystyle x^2\cos x+2 x\sin x$
\item[12.] $\displaystyle \left(2x+3\right)\cot\left(x^2+3x\right)$ \ \ \ \ \ Recall $\displaystyle\cot\theta=\frac{\cos\theta}{\sin\theta}$
\item[14.] $\displaystyle 2\log x\frac{1}{x\ln 10}$
\item[16.] $\displaystyle 2\csc(x-y)+1$
\item[18.] $\displaystyle \frac{1}{x+10}$
\item[20.] $\displaystyle \frac{2x+3}{x^2+3x}$
\item[22.] $\displaystyle 2\sec^2x\tan x$
\item[24.] $\displaystyle \arcsin 2x+\frac{2x}{\sqrt{1-4x^2}}$
\end{description}
\clearpage
\section*{Basic Formulas}
\setcounter{equation}{0}

\textbf{Derivatives of Trigonometric Functions}
\begin{boxedeqnarray2}
\frac{d}{dx}\left(\sin x\right) & = & \cos x \\
\frac{d}{dx}\left(\cos x\right) & = & -\sin x \\
\frac{d}{dx}\left(\tan x\right) & = & \sec^2 x \\
\frac{d}{dx}\left(\cot x\right) & = & -\csc^2 x \\
\frac{d}{dx}\left(\sec x\right) & = & \sec x\tan x \\
\frac{d}{dx}\left(\csc x\right) & = & -\csc x\cot x
\end{boxedeqnarray2}

\textbf{Derivatives of Inverse Trigonometric Functions}
\begin{boxedeqnarray2}
\frac{d}{dx}\left(\sin^{-1} x\right) & = & \frac{1}{\sqrt{1-x^2}} \\
\frac{d}{dx}\left(\cos^{-1} x\right) & = & \frac{-1}{\sqrt{1-x^2}} \\
\frac{d}{dx}\left(\tan^{-1} x\right) & = & \frac{1}{1+x^2} \\
\frac{d}{dx}\left(\cot^{-1} x\right) & = & \frac{-1}{1+x^2} \\
\frac{d}{dx}\left(\sec^{-1} x\right) & = & \frac{1}{|x|\sqrt{x^2-1}} \\
\frac{d}{dx}\left(\csc^{-1} x\right) & = & \frac{-1}{|x|\sqrt{x^2-1}}
\end{boxedeqnarray2}

\textbf{Derivatives of Logarithmic Functions}
\begin{boxedeqnarray2}
\frac{d}{dx}\left(\ln x\right) & = & \frac{1}{x} \\
\frac{d}{dx}\left(\log_b x\right) & = & \frac{1}{x\ln b}
\end{boxedeqnarray2}

\textbf{Derivatives of Exponential Functions}
\begin{boxedeqnarray2}
\frac{d}{dx}\left(e^x\right) & = & e^x \\
\frac{d}{dx}\left(b^x\right) & = & b^x\ln b
\end{boxedeqnarray2}
\clearpage
\section*{Chain Rule Formulas}
\setcounter{equation}{0}

\textbf{Derivatives of Trigonometric Functions}
\begin{boxedeqnarray2}
\frac{d}{dx}\left(\sin u\right) & = & \cos u \frac{du}{dx} \\
\frac{d}{dx}\left(\cos u\right) & = & -\sin u \frac{du}{dx} \\
\frac{d}{dx}\left(\tan u\right) & = & \sec^2 u \frac{du}{dx} \\
\frac{d}{dx}\left(\cot u\right) & = & -\csc^2 u \frac{du}{dx} \\
\frac{d}{dx}\left(\sec u\right) & = & \sec u\tan u \frac{du}{dx} \\
\frac{d}{dx}\left(\csc u\right) & = & -\csc u\cot u \frac{du}{dx}
\end{boxedeqnarray2}

\textbf{Derivatives of Inverse Trigonometric Functions}
\begin{boxedeqnarray2}
\frac{d}{dx}\left(\sin^{-1} u\right) & = & \frac{1}{\sqrt{1-u^2}} \frac{du}{dx} \\
\frac{d}{dx}\left(\cos^{-1} u\right) & = & \frac{-1}{\sqrt{1-u^2}} \frac{du}{dx} \\
\frac{d}{dx}\left(\tan^{-1} u\right) & = & \frac{1}{1+u^2} \frac{du}{dx} \\
\frac{d}{dx}\left(\cot^{-1} u\right) & = & \frac{-1}{1+u^2} \frac{du}{dx} \\
\frac{d}{dx}\left(\sec^{-1} u\right) & = & \frac{1}{|u|\sqrt{u^2-1}} \frac{du}{dx} \\
\frac{d}{dx}\left(\csc^{-1} u\right) & = & \frac{-1}{|u|\sqrt{u^2-1}} \frac{du}{dx}
\end{boxedeqnarray2}

\textbf{Derivatives of Logarithmic Functions}
\begin{boxedeqnarray2}
\frac{d}{dx}\left(\ln u\right) & = & \frac{1}{u} \frac{du}{dx} \\
\frac{d}{dx}\left(\log_b u\right) & = & \frac{1}{u\ln b} \frac{du}{dx}
\end{boxedeqnarray2}

\textbf{Derivatives of Exponential Functions}
\begin{boxedeqnarray2}
\frac{d}{dx}\left(e^u\right) & = & e^u \frac{du}{dx} \\
\frac{d}{dx}\left(b^u\right) & = & b^u\ln b \; \frac{du}{dx}
\end{boxedeqnarray2}

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