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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


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

\maketitle

\subsection{Differentiation Rules}

\begin{example}
Find \[
\frac{d}{dx}\left( \left( x^{2}+1\right) \left( x^{3}-2x\right) \left( 3x^{4}+2x^{2}+2\right) \right) .\]

\end{example}

\subsubsection*{Solution}

To use the product rule correctly, you are supposed to only have a product of
two things. We can group together two of these three and use the product rule
first on them:

\begin{eqnarray*}
 &  & \frac{d}{dx}\left[ \left( x^{2}+1\right) \left( x^{3}-2x\right) \left( 3x^{4}+2x^{2}+2\right) \right] \\
 & = & \frac{d}{dx}\left[ \left( \left( x^{2}+1\right) \left( x^{3}-2x\right) \right) \left( 3x^{4}+2x^{2}+2\right) \right] \\
 & = & \left[ \frac{d}{dx}\left( \left( x^{2}+1\right) \left( x^{3}-2x\right) \right) \right] \left( 3x^{4}+2x^{2}+2\right) +\left( \left( x^{2}+1\right) \left( x^{3}-2x\right) \right) \frac{d}{dx}\left( 3x^{4}+2x^{2}+2\right) \\
 & = & \left[ \left( \frac{d}{dx}\left( x^{2}+1\right) \right) \left( x^{3}-2x\right) +\left( x^{2}+1\right) \frac{d}{dx}\left( x^{3}-2x\right) \right] \left( 3x^{4}+2x^{2}+2\right) \\
 &  & \quad +\left( x^{2}+1\right) \left( x^{3}-2x\right) \frac{d}{dx}\left( 3x^{4}+2x^{2}+2\right) \\
 & = & \left( \frac{d}{dx}\left( x^{2}+1\right) \right) \left( x^{3}-2x\right) \left( 3x^{4}+2x^{2}+2\right) +\left( x^{2}+1\right) \left( \frac{d}{dx}\left( x^{3}-2x\right) \right) \left( 3x^{4}+2x^{2}+2\right) \\
 &  & \quad +\left( x^{2}+1\right) \left( x^{3}-2x\right) \frac{d}{dx}\left( 3x^{4}+2x^{2}+2\right) .
\end{eqnarray*}
Boy, that sure stretches across the screen. But look at what you have there.
The derivative of the first, times the next two, then the first times the derivative
of the second times the third, then the first two times the derivative of the
third. This works in general:\[
\left( f(x)g(x)h(x)\right) '=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x).\]
It also works for the derivative of products of more factors. You have a new
term for each factor, each one with one of the factors differentiated. 

To finish off this example, finally I am going to differentiate the separate
terms:

\begin{eqnarray*}
 &  & \frac{d}{dx}\left[ \left( x^{2}+1\right) \left( x^{3}-2x\right) \left( 3x^{4}+2x^{2}+2\right) \right] \\
 & = & \left( 2x\right) \left( x^{3}-2x\right) \left( 3x^{4}+2x^{2}+2\right) +\left( x^{2}+1\right) \left( 3x^{2}-2\right) \left( 3x^{4}+2x^{2}+2\right) \\
 &  & \quad +\left( x^{2}+1\right) \left( x^{3}-2x\right) \left( 12x^{3}+4x\right) .
\end{eqnarray*}

\end{document}
