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%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
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\section{ Tables of Differentiation Rules and Formulas\label%
{table of differentiation formulas}\qquad}

The \label{derivative}\textbf{derivative of a function $f$ at a number }$x$,
denoted by $f^{\ \prime }(x)$, $\dfrac{d}{dx}f\left( x\right) $, $\dfrac{df}{%
dx}\left( x\right) $, or $D_{x}\left( f(x)\right) $ is the limit 
\[
f^{\ \prime }(x)=\lim_{h\stackrel{}{\rightarrow }0}\,\,\frac{f(x+h)-f(x)}{h} 
\]
if this limit exists.\bigskip

This associates with a function $y=f(x)$, a new function $\frac{dy}{dx}=f^{\
\prime }(x)$, called the \textbf{derivative of }$f$.\medskip

\subsection{ Differentiation Rules\label{differentiation rules}}

In this table, $f$, $g$, $u$, $v$, and $y$ represent functions of $x$, and $%
c $ and $n$ represent constants.\medskip

$
\begin{array}{llllll}
\text{\textsl{1.}} & c^{\prime }=0 & \qquad & \dfrac{d}{dx}\left( c\right) =0
& \qquad &  \\ 
&  &  &  &  &  \\ 
\text{\textsl{2.}} & (cf\,)^{\prime }=cf^{\ \prime } &  & \dfrac{d}{dx}%
\left( cu\right) =c\dfrac{du}{dx} &  &  \\ 
&  &  &  &  &  \\ 
\text{\textsl{3.}} & (f+g)^{\prime }=f^{\ \prime }+g^{\prime } &  & \dfrac{d%
}{dx}\left( u+v\right) =\dfrac{du}{dx}+\dfrac{dv}{dx} &  & \text{\label%
{sum rule}(sum rule)} \\ 
&  &  &  &  &  \\ 
\text{\textsl{4.}} & (f-g)^{\prime }=f^{\ \prime }-g^{\prime } &  & \dfrac{d%
}{dx}\left( u-v\right) \left( x\right) =\dfrac{du}{dx}-\dfrac{dv}{dx} &  & 
\\ 
&  &  &  &  &  \\ 
\text{\textsl{5.}} & \,(x^{n})^{\prime }=nx^{n-1} &  & \dfrac{d}{dx}\left(
x^{n}\right) =nx^{n-1} &  & \text{(power rule)\label{power rule}} \\ 
&  &  &  &  &  \\ 
\text{\textsl{6.}} & (f\left( x\right) ^{n})^{\prime }= &  & \dfrac{d}{dx}%
\left( u^{n}\right) =nu^{n-1}\dfrac{du}{dx} &  & \text{(generalized power
rule)} \\ 
&  &  &  &  &  \\ 
\text{\textsl{7.}} & (fg)^{\prime }=f^{\ \prime }g+fg^{\prime } &  & \dfrac{d%
}{dx}\left( uv\right) =u\dfrac{dv}{dx}+v\dfrac{du}{dx} &  & \text{(product
rule)\label{product rule}} \\ 
&  &  &  &  &  \\ 
\text{\textsl{8.}} & \left( \dfrac{f}{g}\right) ^{\prime }=\dfrac{f^{\
\prime }g-fg^{\prime }}{g^{2}} &  & \dfrac{d}{dx}\left( \dfrac{u}{v}\right) =%
\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^{2}} &  & \text{(quotient rule)%
\label{quotient rule}} \\ 
&  &  &  &  &  \\ 
\text{\textsl{9.}} & \left( f\circ g\right) ^{\prime }=\left( f^{\ \prime
}\circ g\right) g^{\prime } &  & \dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}
&  & \text{(chain rule)\label{chain rule}}
\end{array}
\medskip $

\subsection{ Differentiation Formulas\label{differentiation formulas}}

In these formulas, $u$ and $v$ represent functions of $x$; $a$, $b$ and $n$
represent constants; $e=\allowbreak 2.\,\allowbreak 71828....$ and $\pi
=\allowbreak 3.\,\allowbreak 14159...$ ; and all angles are measured in
radians.\medskip

\subsubsection{\label{algebraic functions}Algebraic Functions}

\begin{enumerate}
\item  $\dfrac{d}{dx}\left( au\pm bv\right) =a\dfrac{du}{dx}\pm b\dfrac{dv}{%
dx}$

\item  $\dfrac{d}{dx}\left( u^{n}\right) =nu^{n-1}\dfrac{du}{dx}$
\end{enumerate}

\subsubsection{\label{trigonometric functions}Trigonometric Functions}

\begin{enumerate}
\item  $\dfrac{d}{dx}\left( \sin u\right) =\cos u\dfrac{du}{dx}$

\item  $\dfrac{d}{dx}\left( \cos u\right) =-\sin u\dfrac{du}{dx}$

\item  $\dfrac{d}{dx}\left( \tan u\right) =\allowbreak \sec ^{2}u\dfrac{du}{%
dx}$

\item  $\dfrac{d}{dx}\left( \cot u\right) =\allowbreak -\csc ^{2}u\dfrac{du}{%
dx}$

\item  $\dfrac{d}{dx}\left( \sec u\right) =\allowbreak \sec u\tan u\dfrac{du%
}{dx}$

\item  $\dfrac{d}{dx}\left( \csc u\right) =\allowbreak -\csc u\cot u\dfrac{du%
}{dx}$

\item  $\dfrac{d}{dx}\left( \limfunc{vers}u\right) =\sin u\dfrac{du}{dx}$
\end{enumerate}

\subsubsection{\label{inverse trigonometric functions}Inverse Trigonometric
Functions}

\begin{enumerate}
\item  $\dfrac{d}{dx}\left( \sin ^{-1}u\right) =\frac{1}{\sqrt{1-u^{2}}}%
\dfrac{du}{dx}$

\item  $\dfrac{d}{dx}\left( \cos ^{-1}u\right) =-\frac{1}{\sqrt{1-u^{2}}}%
\dfrac{du}{dx}$

\item  $\dfrac{d}{dx}\left( \tan ^{-1}u\right) =\frac{1}{1+u^{2}}\dfrac{du}{%
dx}$

\item  $\dfrac{d}{dx}\left( \cot ^{-1}u\right) =-\frac{1}{1+u^{2}}\dfrac{du}{%
dx}$

\item  $\dfrac{d}{dx}\left( \sec ^{-1}u\right) =\frac{1}{u\sqrt{u^{2}-1}}%
\dfrac{du}{dx}$, $-\pi \leq \sec ^{-1}u<-\frac{\pi }{2}$, $0\leq \sec ^{-1}u<%
\frac{\pi }{2}$

\item  $\dfrac{d}{dx}\left( \csc ^{-1}u\right) =-\frac{1}{u\sqrt{u^{2}-1}}%
\dfrac{du}{dx}$, $-\pi <\csc ^{-1}u\leq -\frac{\pi }{2}$, $0<\sec ^{-1}u\leq 
\frac{\pi }{2}$
\end{enumerate}

\subsubsection{\label{exponential and logarithmic functions}Exponential and
Logarithmic Functions}

\begin{enumerate}
\item  $\dfrac{d}{dx}\left( e^{u}\right) =e^{u}\dfrac{du}{dx}$

\item  $\dfrac{d}{dx}\left( a^{u}\right) =a^{u}\log _{e}a\dfrac{du}{dx}$

\item  $\dfrac{d}{dx}\left( \ln u\right) =\dfrac{1}{u}\dfrac{du}{dx}$

\item  $\dfrac{d}{dx}\left( \log _{a}u\right) =\frac{\log _{a}e}{u}\dfrac{du%
}{dx}$

\item  $\dfrac{d}{dx}\left( u^{v}\right) =vu^{v-1}\dfrac{du}{dx}+u^{v}\ln u%
\dfrac{dv}{dx}$
\end{enumerate}

\subsubsection{\label{hyperbolic functions}Hyperbolic Functions}

\begin{enumerate}
\item  $\dfrac{d}{dx}\left( \sinh u\right) =\cosh u\dfrac{du}{dx}$

\item  $\dfrac{d}{dx}\left( \cosh u\right) =\sinh u\dfrac{du}{dx}$

\item  $\dfrac{d}{dx}\left( \tanh u\right) =\func{sech}^{2}u\dfrac{du}{dx}$

\item  $\dfrac{d}{dx}\left( \coth u\right) =-\func{csch}^{2}u\dfrac{du}{dx}$

\item  $\dfrac{d}{dx}\left( \func{sech}u\right) =-\func{sech}u\tanh u\dfrac{%
du}{dx}$

\item  $\dfrac{d}{dx}\left( \func{csch}u\right) =-\func{csch}u\coth u\dfrac{%
du}{dx}$
\end{enumerate}

\subsubsection{Inverse Hyperbolic Functions\label%
{inverse hyperbolic functions}}

\begin{enumerate}
\item  $\dfrac{d}{dx}\left( \sinh ^{-1}u\right) =\frac{1}{\sqrt{u^{2}+1}}%
\dfrac{du}{dx}$

\item  $\dfrac{d}{dx}\left( \cosh ^{-1}u\right) =\frac{1}{\sqrt{u^{2}-1}}%
\dfrac{du}{dx}$, $u>1$

\item  $\dfrac{d}{dx}\left( \tanh ^{-1}u\right) =\frac{1}{1-u^{2}}\dfrac{du}{%
dx}$

\item  $\dfrac{d}{dx}\left( \coth ^{-1}u\right) =\frac{1}{1-u^{2}}\dfrac{du}{%
dx}$

\item  $\dfrac{d}{dx}\left( \func{sech}^{-1}x\right) =-\allowbreak \frac{1}{u%
\sqrt{1-u^{2}}}\dfrac{du}{dx}$, $u>0$

\item  $\dfrac{d}{dx}\left( \func{csch}^{-1}u\right) =\allowbreak -\frac{1}{u%
\sqrt{1+u^{2}}}\dfrac{du}{dx}$
\end{enumerate}

\strut

\subsection{\ Related Reference Topics}

{\small \hyperref{Calculus}{}{}{cal_contents.tex}}

{\small \hyperref{Table of Integrals}{}{}{TAB_integrals.tex}}

\subsection{\ Related Scientific Notebook Topics\hfill}

{\small \hyperref{Differentiation}{}{}{../../Help/DM7-3.tex#Derivative}}

\rule{0.67in}{0.01in}

\quad {\small Math Index entries: \hyperref{Calculus}{}{}{%
../../Help/DMIndex.tex#Calculus}, \hyperref{Derivative}{}{}{%
../../Help/DMIndex.tex#Derivative}, \hyperref{Directional derivative}{}{}{%
../../Help/DMIndex.tex#Directional derivative}}

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