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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsubsection{Derivatives of inverse hyperbolic functions}

If this isn't enough, there are inverses, too. However, from the point of view
of calculus, even the trigonometric inverses weren't all that interesting, except
that the derivatives were more basic functions. That will be useful when it
comes time to find integals of those expressions. 

So, inverse hyperbolic functions shouldn't be interesting (in calculus) either,
unless they provide new integration formulas. However, they don't (at least,
none that can't be easily found other ways). However, there is an odd collection
of results: these inverses can be written easily in terms of functions we already
know, such as: 
\[
\sinh ^{-1}x=\ln (x+\sqrt{x^{2}+1}),\]
 which follows from the definitions. 

Why is that true? Well, set \( y=\sinh ^{-1}(x) \), so 
\begin{eqnarray*}
x & = & \sinh (y)\\
 & = & \frac{1}{2}\left( e^{y}-e^{-y}\right) ,
\end{eqnarray*}
or, solving for \( e^{y} \) by first multiplying both sides by \( e^{y} \),
which gives a quadratic in \( e^{y} \),
\begin{eqnarray*}
2e^{y}x & = & \left( e^{y}\right) ^{2}-1,\\
0 & = & \left( e^{y}\right) ^{2}-2e^{y}x-1,
\end{eqnarray*}
which by the quadratic formula gives
\begin{eqnarray*}
e^{y} & = & \frac{2x\pm \sqrt{4x^{2}+4}}{2}\\
 & = & x\pm \sqrt{x^{2}+1}.
\end{eqnarray*}
However, the ``\( - \)'' of the \( \pm  \) can't possibly be correct, since
that would be a negative number on the right (why? And why is that a problem?),
so 
\[
e^{y}=x+\sqrt{x^{2}+1}.\]
Taking natural logarithms of both sides gives
\[
y=\ln \left( x+\sqrt{x^{2}+1}\right) .\]


Now, finding derivatives of \( \sinh ^{-1}x \) is just an exercise in the chain
rule:
\begin{eqnarray*}
\left( \sinh ^{-1}x\right) ' & = & \left( \ln \left( x+\sqrt{x^{2}+1}\right) \right) '\\
 & = & \frac{1}{\left( x+\sqrt{x^{2}+1}\right) }\left( 1+\frac{x}{\sqrt{x^{2}+1}}\right) \\
 & = & \frac{1}{\left( x+\sqrt{x^{2}+1}\right) }\left( \frac{\sqrt{x^{2}+1}+x}{\sqrt{x^{2}+1}}\right) \\
 & = & \frac{1}{\sqrt{x^{2}+1}}.
\end{eqnarray*}
There are similar formulas for the other inverse hyperbolic functions, some
of which are derived in the examples.

\end{document}
