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	{\noindent 21-120 Differential and Integral Calculus \hfill Summer Session I 2010\\}
	{\centering Quiz 6\medskip\\}
	\hrule
	{
		\mbox{}\\
		\begin{prob*} \mbox{}
		\begin{enumerate}
			\item Find $y'$ if \[ y=\ln(e^{2x+3})\cdot\cos(e^{2x}) \]
			\item Solve \[ 2^{x-5} = 4 \]
		\end{enumerate}
		\end{prob*}
		\begin{soln*}
			\mbox{}
			\begin{enumerate}
				\item Note that, $\ln(e^{2x+3}) = 2x+3$. Thus we will need to take the derivative of $y=(2x+3)\cdot\cos(e^{2x})$. We use the product rule.
				\[ \frac{dy}{dx}=\frac{d}{dx}\left((2x+3)\cdot\cos(e^{2x})\right) = \frac{d}{dx}\left(2x+3\right)\cdot\cos(e^{2x}) + (2x+3)\cdot\frac{d}{dx}\left(\cos(e^{2x})\right)\]
				$ \frac{d}{dx}\left(2x+3\right) = 2$ simply by the Power Rule. For $\frac{d}{dx}\left(\cos(e^{2x}\right)$ we use the chain rule repeatedly:
				\begin{align*}
					\frac{d}{dx}\left(\cos(e^{2x})\right) &= -\sin(e^{2x})\cdot \frac{d}{dx}\left(e^{2x}\right) \\
					&= -\sin(e^{2x}) \cdot e^{2x} \cdot \frac{d}{dx}\left(2x\right) \\
					&=
				\end{align*}
				Thus, we have that
				\[ y'=\frac{dy}{dx} = 2\cos(e^{2x}) -2e^{2x}(2x+3)\sin(e^{2x}) \]
				\item \mbox{}
				\begin{align*}
					 & 2^{x-5} = 4 \\
					 \implies & \log_2(2^{x-5}) = \log_2(4) \\
					 \implies & x-5 = \log_2(4)=\log_2(2^2) = 2\\
					 \implies & x=7
				\end{align*}
				Alternatively,

				  \begin{align*}
				  	& 2^{x-5} = 2^2 \\
				  	\implies & x-5 = 2 \\
				  	\implies & x=7
				  \end{align*}

			\end{enumerate}

		\end{soln*}

	}
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