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\noindent \hfill
\fbox{\fbox{\parbox{7.25in}{\scshape{Math 1110 \hfill Derivatives Practice}}}} \hfill \\

\begin{equation*} \begin{array}{rlrl} \vspace{0.05in}
 1. & f(x) = 3x^2-\ln(x)+e^x+5x-2 & \qquad \qquad 2. & f(x) = \sqrt{x} + \frac{2}{x} -3x^5 \\ \vspace{0.05in}
 3. & y = x^3e^x                  & \qquad \qquad 4. & y = \frac{x^2+1}{x^3-2x+1}          \\ \vspace{0.05in}
 5. & y = (\ln(x)+2x-7)^4         & \qquad \qquad 6. & f(x) = \ln\left(\frac{x^2-2}{e^{3x}}\right) \\ \vspace{0.05in}
 7. & y = \ln(4x^3(x^2+1))        & \qquad \qquad 8. & f(x) = \frac{\sqrt{x}}{x^3}+e^{-2x+\ln(3x)}  \\ \vspace{0.05in}
 9. & f(x) = \ln(\sqrt{x^3-x}+e^{5x}) & \qquad \qquad 10. & y = e^{e^{e^x}}                 \\ \vspace{0.05in}
11. & y = x^3\ln(x+3)e^{-x}       & \qquad \qquad 12. & f(x) = \frac{\ln(3x+2)}{e^x+x}     \\ \vspace{0.05in}
13. & y = \ln(\sqrt{x}e^{x^2-2x}) & \qquad \qquad 14. & f(x) = (\ln(e^x+1))^4              \\ \vspace{0.05in}
\end{array} \end{equation*}

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\force \qquad \qquad
\noindent
Find an equation for the line tangent to $y=f(x)$ at the given point.
\begin{equation*} \begin{array}{lclc} \vspace{0.05in}
y=x^2 \quad \text{at} \quad x=2 & \qquad \qquad \qquad & y=\ln(x) \quad \text{at} \quad x=1 & \qquad \qquad \\ \vspace{0.05in}
y=e^x \quad \text{at} \quad x=0 & \qquad \qquad \qquad & y=\sqrt{x} \quad \text{at} \quad \qquad x=9 & \qquad \qquad  
\end{array}\end{equation*}

\vspace{0.1in}

{\bf\Large Answers:}


\begin{enumerate}
\item $\displaystyle{f(x) = 3x^2-\ln(x)+e^x+5x-2}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{f'(x)=6x-\frac{1}{x}+e^x+5}$

\item $\displaystyle{f(x) = \sqrt{x} + \frac{2}{x} -3x^5}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{f(x) = x^{1/2} +2x^{-1}-3x^5}$ \quad $\Longrightarrow$ \quad\\
      $\displaystyle{f'(x) = \frac{1}{2}x^{-1/2}-2x^{-2}-15x^4}$

\item $\displaystyle{y = x^3e^x}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{y' = 3x^2e^x+x^3e^x}$

\item $\displaystyle{y = \frac{x^2+1}{x^3-2x+1}}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{y'=\frac{2x(x^3-2x+1)-(x^2+1)(3x^2-2)}{(x^3-2x+1)^2}}$ \quad $\Longrightarrow$ \quad\\
      $\displaystyle{y'=\frac{2x^4-4x^2+2x-3x^4-3x^2+2x^2+2}{(x^3-2x+1)^2}}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{y'=\frac{-x^4-5x^2+2x+2}{(x^3-2x+1)^2}}$

\item $\displaystyle{y = (\ln(x)+2x-7)^4}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{y' = 4(\ln(x)+2x-7)^3\left(\frac{1}{x}+2\right)}$

\item $\displaystyle{f(x) = \ln\left(\frac{x^2-2}{e^{3x}}\right)}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{f(x) = \ln(x^2-2)-\ln(e^{3x})}$ \quad $\Longrightarrow$ \quad\\
      $\displaystyle{f(x) = \ln(x^2-2)-3x}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{f'(x) = \frac{2x}{x^2-2}-3}$

\item $\displaystyle{y = \ln(4x^3(x^2+1))}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{y = \ln(4)+\ln(x^3)+\ln(x^2+1)}$ \quad $\Longrightarrow$ \quad\\
      $\displaystyle{y = \ln(4)+3\ln(x)+\ln(x^2+1)}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{y' = \frac{3}{x}+\frac{2x}{x^2+1}}$

\item $\displaystyle{f(x) = \frac{\sqrt{x}}{x^3}+e^{-2x+\ln(3x)}}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{f(x) = x^{1/2}x^{-3}+e^{-2x}e^{\ln(3x)}}$ \quad $\Longrightarrow$ \quad\\
      $\displaystyle{f(x) = x^{-5/2}+e^{-2x}3x}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{f'(x) = -\frac{5}{2}x^{-7/2}+e^{-2x}(-2)3x+e^{-2x}3}$ \quad $\Longrightarrow$ \quad\\
      $\displaystyle{f'(x) = -\frac{5}{2}x^{-7/2}-6xe^{-2x}+3e^{-2x}}$

\item $\displaystyle{f(x) = \ln(\sqrt{x^3-x}+e^{5x})}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{f(x) = \ln((x^3-x)^{1/2}+e^{5x})}$ \quad $\Longrightarrow$ \quad\\
      $\displaystyle{f'(x) = \frac{1}{(x^3-x)^{1/2}+e^{5x}}\left(\frac{1}{2}(x^3-x)^{-1/2}(3x^2-1)+e^{5x}5\right)}$

\item $\displaystyle{y = e^{e^{e^x}}}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{y' = e^{e^{e^x}}e^{e^x}e^x}$

\item $\displaystyle{y = x^3\ln(x+3)e^{-x}}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{y'=3x^2\ln(x+3)e^{-x}+x^3\frac{1}{x+3}e^{-x}+x^3\ln(x+3)e^{-x}(-1)}$ 

\item $\displaystyle{f(x) = \frac{\ln(3x+2)}{e^x+x}}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{f'(x) = \frac{\frac{3}{3x+2}(e^x+x)-\ln(3x+2)(e^x+1)}{(e^x+x)^2}}$

\item $\displaystyle{y = \ln(\sqrt{x}e^{x^2-2x})}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{y = \ln(x^{1/2})+\ln(e^{x^2-2x})}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{y = \frac{1}{2}\ln(x)+x^2-2x}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{y' = \frac{1}{2x}+2x-2}$

\item $\displaystyle{f(x) = (\ln(e^x+1))^4}$ \quad $\Longrightarrow$ \quad
      $\displaystyle{f'(x) = 4(\ln(e^x+1))^3\frac{1}{e^x+1}e^x}$
\end{enumerate}

\vspace{0.2in}

\noindent
Find an equation for the line tangent to $y=f(x)$ at the given point.
\begin{itemize}
\item $y=x^2$ at $x=2$.\\
      Notice that $y'=2x$.
      The slope of the tangent of $y=x^2$ at $x=2$ is\\ 
      $\displaystyle{\left. y' \right|_{x=2} = \left. 2x \right|_{x=2} = 2(2) = 4}$.
      If $x=2$, then $y=2^2=4$.\\
      Using point slope: $y-4=4(x-2)$.\\
      The equation of the tangent is $y=4x-4$.
\item $y=\ln(x)$ at $x=1$.\\
      Notice that $y'=1/x$.
      The slope of the tangent of $y=\ln(x)$ at $x=1$ is\\ 
  $\displaystyle{\left. y' \right|_{x=1} = \left. \frac{1}{x} \right|_{x=1} = \frac{1}{1} = 1}$.
      If $x=1$, then $y=\ln(1)=0$.\\
      Using point slope: $y-0=1(x-1)$.\\
      The equation of the tangent is $y=x-1$.
\item $y=e^x$ at $x=0$.\\
      Notice that $y'=e^x$.
      The slope of the tangent of $y=e^x$ at $x=0$ is\\
  $\displaystyle{\left. y' \right|_{x=0} = \left. e^x \right|_{x=0} = e^0 = 1}$.\\
      If $x=0$, then $y=e^0=1$. Using point slope: $y-1=1(x-0)$.
      The equation of the tangent is $y=x+1$.
\item $y=\sqrt{x}=x^{1/2}$ at $x=9$.\\
      Notice that $y'=(1/2)x^{-1/2}$.
      The slope of the tangent of $y=x^{1/2}$ at $x=9$ is 
  $\displaystyle{\left. y' \right|_{x=9} = \left. (1/2)x^{-1/2} \right|_{x=9} = \frac{1}{2\sqrt{9}} = \frac{1}{6}}$.\\
      If $x=9$, then $y=\sqrt{9}=3$.\\
      Using point slope: $y-3=(1/6)(x-3)$.\\
      The equation of the tangent is $\displaystyle{y=\frac{1}{6}x+\frac{3}{2}}$.
\end{itemize}

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