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\problemgrouptitle{Definition of Derivative}

\begin{problemblock}{Use the definition of the derivative at a value $a$ to find the following.}

\item $f'(2)$ if $\displaystyle f(x)=\frac{1}{x+1}$
\answer{$\displaystyle f'(2)=\lim_{h\to 0}\frac{\frac{1}{(2+h)+1}-\frac{1}{2+1}}{h}=-\frac{1}{9}$}

\item $g'(4)$ if $\displaystyle g(x)=\sqrt{2x}$
\answer{$\displaystyle g'(4)=\lim_{h\to 0}\frac{\sqrt{2(4+h)}-\sqrt{2(4)}}{h}=\frac{1}{2\sqrt{2}}$}

\end{problemblock}

\begin{problemblock}[2]{Use the definition of the derivative to calculate $f'(x)$ for each of the following functions.}

\item $\displaystyle f(x)=x^2+3$ 
\answer{$\displaystyle f'(x)=\lim_{h\to 0}\frac{(x+h)^2+3-(x^2+3)}{h}=2x$}

\item $f(x)=\displaystyle\frac{1}{3x}$
\answer{$\displaystyle f'(x)=\lim_{h\to 0}\frac{\frac{1}{3(x+h)}-\frac{1}{3x}}{h}=-\frac{1}{3x^2}$}

\item $f(x)=(x+2)^2$
\answer{$\displaystyle f'(x)=\lim_{h\to 0}\frac{(x+2+h)^2-(x+2)^2}{h}=2x+4$}

\item $\displaystyle f(x)=\sqrt{x+2}$
\answer{$\displaystyle f'(x)=\lim_{h\to 0}\frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}=\frac{1}{2\sqrt{x+2}}$}


\item $\displaystyle f(x)=\frac{3x+2}{x+1}$ 
\answer{$\displaystyle f'(x)=\lim_{h\to 0}\frac{\frac{3(x+h)+2}{x+h+1}-\frac{3x+2}{x+1}}{h}=\frac{1}{(x+1)^2}$}

\item $\displaystyle f(x)=\frac{1}{\sqrt{x}}$
\answer{$\displaystyle f'(x)=\lim_{h\to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}=-\frac{1}{2x\sqrt{x}}=-\frac{1}{2}x^{-\frac{3}{2}}$}

\end{problemblock}

\item Prove that $f(x)=\sqrt{(x-2)^2}$ is not differentiable at $x=2$ by showing that the following left and right hand limits differ:

$\displaystyle\lim_{h\to0^{-}}\frac{f(2+h)-f(2)}{h}\neq \lim_{h\to0^{+}}\frac{f(2+h)-f(2)}{h}$.
\answer{$\displaystyle\lim_{h\to 0^-}\frac{\displaystyle\sqrt{(2+h-2)^2}}{h}=\lim_{h\to 0^-}\frac{\sqrt{h^2}}{h}=\lim_{h\to 0^-}\frac{|h|}{h}=\lim_{h\to 0^-}\frac{-h}{h}=-1$

$\displaystyle\lim_{h\to 0^+}\frac{\displaystyle\sqrt{(2+h-2)^2}}{h}=\lim_{h\to 0^+}\frac{\sqrt{h^2}}{h}=\lim_{h\to 0^+}\frac{|h|}{h}=\lim_{h\to 0^+}\frac{h}{h}=1$}
