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% These are all implicit differentiation problems
\newproblem{imd:circ}{%
Find the gradient of the unit circle ($x^2 + y^2 = 1$).}{%
Differentiating with respect to $x$ gives:
\begin{eqnarray*}
2x + 2y\frac{dy}{dx} & = & 0\\
\frac{dy}{dx} & = & \frac{-2x}{2y}\\
& = & \frac{-x}{\sqrt{1-x^2}}.
\end{eqnarray*}}
\newproblem{imd:ysq:xcuov2mx}{%
Find $\frac{dy}{dx}$, given
\begin{displaymath}
y^2 = \frac{x^3}{2-x}
\end{displaymath}}{%
Differentiating both sides w.r.t.\ $x$:
\begin{eqnarray*}
2y\frac{dy}{dx} & = & \frac{(2-x)3x^2 - x^3(-1)}{(2-x)^2}\\
& = & \frac{3x^2(2-x) + x^3}{(2-x)^2}\\
& = & \frac{6x^2 - 3x^3 + x^3}{(2-x)^2}\\
& = & \frac{6x^2-2x^3}{(2-x)^2}\\
& = & 2x^2\frac{3-x}{(2-x)^2}
\end{eqnarray*}
Therefore
\begin{displaymath}
y\frac{dy}{dx} = x^2\frac{3-x}{(2-x)^2}
\end{displaymath}}
\newproblem{imd:exy:IIxay}{%
Differentiate w.r.t.\ $x$:
\begin{displaymath}
e^{xy} = 2x + y
\end{displaymath}}{%
Differentiating both sides w.r.t.\ $x$:
\begin{eqnarray*}
e^{xy}(1y + x\frac{dy}{dx}) & = & 2 + \frac{dy}{dx}\\
xe^{xy}\frac{dy}{dx} - \frac{dy}{dx} & = & 2 - ye^{xy} \\
\frac{dy}{dx}(xe^{xy}-1) & = & 2 - ye^{xy}\\
\frac{dy}{dx} & = & \frac{2-ye^{xy}}{xe^{xy}-1}
\end{eqnarray*}}
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