M-143-Chapter11Help.tex

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\begin{document}
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{\bf Name:} \hrulefill\hrulefill\hrulefill\\
{\bf M143} \qquad \qquad \\
{\bf Chapter 11 Help}\\ %(look familiar??)\\
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%---------------- End of the document ---------------

What is the derivative of $g(x)=\sqrt{11x}$?  There are two ways to do this.

\begin{enumerate}
\item Using the limit definition:

\begin{eqnarray*}
g'(x)&=&\limit{h}{0}\frac{\sqrt{11(x+h)}-\sqrt{11x}}{h}\\
&=&\limit{h}{0}\frac{\sqrt{11x+11h}-\sqrt{11x}}{h}\\
&=&\limit{h}{0}\frac{\sqrt{11x+11h}-\sqrt{11x}}{h}\frac{\sqrt{11x+11h}+\sqrt{11x}}{\sqrt{11x+11h}-\sqrt{11x}}\\
&=&\limit{h}{0}\frac{11x+11h-11x}{h(\sqrt{11x+11h}+\sqrt{11x})}\\
&=&\limit{h}{0}\frac{11h}{h(\sqrt{11x+11h}+\sqrt{11x})}\\
&=&\limit{h}{0}\frac{11}{(\sqrt{11x+11h}+\sqrt{11x})}\\
&=&\frac{11}{(\sqrt{11x+0}+\sqrt{11x})}\\
&=&\frac{11}{2\sqrt{11x}}
\end{eqnarray*}

\item On the other hand, if you've started reading into Chapter 12 or you know something about derivatives:

\begin{eqnarray*}
g(x)&=&\sqrt{11x}=(11x)^{\frac{1}{2}}\\
g'(x)&=&\frac{1}{2}\cdot(11x)^{-\frac{1}{2}}\cdot\frac{d}{dx}[11x],\ \text{by the chain rule:}\\
&=&\frac{1}{2}\cdot(11x)^{-\frac{1}{2}}\cdot11,\ \text{then we rewrite this a bit}:\\
&=&\frac{1}{2}\frac{1}{\sqrt{11x}}11\\
&=&\frac{11}{2\sqrt{11x}}.
\end{eqnarray*}

\end{enumerate}





















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