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\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} &amp; = &amp; 0\\
\frac{dy}{dx} &amp; = &amp; \frac{-2x}{2y}\\
 &amp; = &amp; \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} &amp; = &amp; \frac{(2-x)3x^2 - x^3(-1)}{(2-x)^2}\\
 &amp; = &amp; \frac{3x^2(2-x) + x^3}{(2-x)^2}\\
 &amp; = &amp; \frac{6x^2 - 3x^3 + x^3}{(2-x)^2}\\
 &amp; = &amp; \frac{6x^2-2x^3}{(2-x)^2}\\
 &amp; = &amp; 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}) &amp; = &amp; 2 + \frac{dy}{dx}\\
xe^{xy}\frac{dy}{dx} - \frac{dy}{dx} &amp; = &amp; 2 - ye^{xy} \\
\frac{dy}{dx}(xe^{xy}-1) &amp; = &amp; 2 - ye^{xy}\\
\frac{dy}{dx} &amp; = &amp; \frac{2-ye^{xy}}{xe^{xy}-1}
\end{eqnarray*}}
\endinput
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%% End of file `implicit.tex'.
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