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	{\noindent 21-120 Differential and Integral Calculus \hfill Summer Session I 2010\\}
	{\centering Quiz 2 \medskip\\}
	\hrule
	{
		\begin{prob*} \mbox{}
			\begin{enumerate}
				\item Use the definition of the derivative to find $\frac{d}{dx}f(x)$ given $f(x)=\sqrt{1+2x}$.
				\item Find the equation of line tangent to $f(x)=x^2+x$ at $x=2$
			\end{enumerate}
		\end{prob*}
		\begin{soln*}
			\mbox{}
			\begin{enumerate}
				\item \begin{align*}
						\frac{d}{dx}f(x)=\lim_{h\to 0} \left(\frac{\sqrt{1+2(x+h)} - \sqrt{1+2x}}{h}\right) &= \lim_{h\to 0} \left(\frac{\sqrt{1+2(x+h)} - \sqrt{1+2x}}{h}\cdot\frac{\sqrt{1+2(x+h)}+\sqrt{1+2x}}{\sqrt{1+2(x+h)}+\sqrt{1+2x}}\right)  \\
						&= \lim_{h\to 0} \left(\frac{ 1+2(x+h) - 1 - 2x}{h(\sqrt{1+2(x+h)}+\sqrt{1+2x})}\right)\\
						&= \lim_{h\to 0} \left(\frac{ 1+2x+2h - 1 - 2x}{h(\sqrt{1+2(x+h)}+\sqrt{1+2x})}\right)\\
						&= \lim_{h\to 0} \left(\frac{ 2h}{h(\sqrt{1+2(x+h)}+\sqrt{1+2x})}\right) \\
						&= \lim_{h\to 0} \left(\frac{ 2}{\sqrt{1+2(x+h)}+\sqrt{1+2x}}\right)\\
						&= \frac{ 2}{\sqrt{1+2x}+\sqrt{1+2x}}\\
						&= \frac{1}{\sqrt{1+2x}}
				      \end{align*}
				\item First we need to take the derivative. We will use the power rule and we determine that \[ f'(x) = 2x+1\] Plugging in $x=2$ we see that $f'(2)=5$. The function value at 2 is given by $f(2)=(2)^2+2=6$. Recall point slope form of a line, ie. if $m$ is the slope, and $(x_1,y_1)$ is a point, then the equation of the line is given by
				\[ y - y_1 = m( x- x_1) \]
				Therefore, we can see the equation of the tangent line is
				\[ y - 6 = 5 ( x - 2 ) \]
				Or, written a bit differently
				\[ y = 5x - 4 \]
			\end{enumerate}
		\end{soln*}
	}
\end{document}