 % These all involve differentiating from 1st principles

\newproblem{dfp:xcube}{%
Differentiate $f(x) = x^3$ with respect to $x$ by first principles.}{%
\begin{eqnarray*}
\frac{dy}{dx} & = & \lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x) - f(x)}{\Delta x}\\
 & = & \lim_{\Delta x\rightarrow 0}\frac{(x+\Delta x)^3-x^3}{\Delta x}\\
 & = & \lim_{\Delta x\rightarrow 0}\frac{(x+\Delta x)(x^2+2x\Delta x+(\Delta x)^2)-x^3}{\Delta x}\\
 & = & \lim_{\Delta x\rightarrow 0}\frac{x^3+3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3-x^3}{\Delta x}\\
 & = & \lim_{\Delta x\rightarrow 0}\frac{3x^2\Delta x+3x(\Delta x)^2+(\Delta x)^3}{\Delta x}\\
 & = & \lim_{\Delta x\rightarrow 0}3x^2+3x\Delta x + (\Delta x)^2\\
 & = & 3x^2
\end{eqnarray*}}

\newproblem{dfp:Ioverxsq}{%
Differentiate $\displaystyle f(x) = \frac{1}{x^2}$ with respect to $x$ by first principles.}{%
\begin{eqnarray*}
\frac{df}{dx} & = & \lim_{\Delta x\rightarrow 0}\frac{\frac{1}{(x+\Delta x)^2}-\frac{1}{x^2}}{\Delta x}\\
 & = & \lim_{\Delta x\rightarrow 0}\frac{\frac{x^2-(x+\Delta x)^2}{x^2(x+\Delta x)^2}}{\Delta x}\\
 & = & \lim_{\Delta x\rightarrow 0}\frac{x^2-(x^2+2x\Delta x+(\Delta x)^2)}{x^2\Delta x(x+\Delta x)^2}\\
 & = & \lim_{\Delta x\rightarrow 0}\frac{-2x\Delta x-(\Delta x)^2}{x^2\Delta x(x+\Delta x)^2}\\
 & = & \lim_{\Delta x\rightarrow 0}\frac{-2x-\Delta x}{x^2(x+\Delta x)^2}\\
 & = & \frac{-2x}{x^2x^2}\\
 & = & -\frac{2}{x^3}
\end{eqnarray*}}

\newproblem{dfp:sqrtx}{%
Differentiate from first principles $f(x) = \surd x$}{%
\begin{eqnarray*}
\frac{df}{dx} & = & \lim_{\Delta x\rightarrow 0}\frac{\sqrt{x+\Delta x}-\surd x}{\Delta x}\\
 & = & \lim_{\Delta x\rightarrow 0}\frac{(\sqrt{x+\Delta x}-\surd x)(\sqrt{x+\delta x}+\surd x)}{\Delta x(\sqrt{x+\Delta x}+\surd x)}\\
 & = & \lim_{\Delta x\rightarrow 0}\frac{x+\Delta x - x}{\Delta x(\sqrt{x+\Delta x}+\surd x)}\\
 & = & \lim_{\Delta x\rightarrow 0}\frac{\Delta x}{\Delta x(\sqrt{x+\Delta x}+\Delta x)}\\
 & = & \lim_{\Delta x\rightarrow 0}\frac{1}{\sqrt{x+\Delta x}+\surd x}\\
 & = & \frac{1}{2\surd x}
\end{eqnarray*}}

\newproblem{dfp:cons}{%
Differentiate from first principles $f(x) = c$ where $c$ is a constant.}{%
\begin{eqnarray*}
\frac{df}{dx} & = & \lim_{\Delta x\rightarrow 0}\frac{c-c}{\Delta x}\\
 & = & \lim_{\Delta x\rightarrow 0}0\\
 & = & 0
\end{eqnarray*}}

\newproblem{dfp:cosx}{%
Given 
\begin{eqnarray*}
\lim_{x \rightarrow 0} \frac{\cos x - 1}{x} & = & 0\\
\lim_{x \rightarrow 0} \frac{\sin x}{x} & = & 1
\end{eqnarray*}
differentiate from first principles $f(x) = \cos x$.}{%
\begin{eqnarray*}
\frac{df}{dx} & = & \lim_{\Delta x \rightarrow 0}\frac{f(x + \Delta x) - f(x)}{\Delta x}\\
 & = & \lim_{\Delta x \rightarrow 0} \frac{\cos(x + \Delta x) - \cos(x)}{\Delta x}\\
 & = & \lim_{\Delta x \rightarrow 0} \frac{\cos x\cos\Delta x - \sin x\sin\Delta x - \cos x}{\Delta x}\\
 & = & \lim_{\Delta x \rightarrow 0} \frac{\cos x(\cos\Delta x - 1) - \sin x\sin\Delta x}{\Delta x}\\
 & = & \cos x\lim_{\Delta x \rightarrow 0}\frac{\cos\Delta x - 1}{\Delta x} 
      - \sin x\lim_{\Delta x \rightarrow 0}\frac{\sin\Delta x}{\Delta x}\\
 & = & -1 \qquad\mbox{(using given results)}
\end{eqnarray*}}
