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%  Differentiation Rules: Implicit Differentiation                 %
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\problemgrouptitle{Differentiation Rules: Implicit Differentiation}

\begin{problemblock}{Calculate $y'$ for functions $y=y(x)$ defined implicitly by the following equations. (Any value that is not $x$ or $y$ should be considered a constant.)}

\item $x^2+y^2=3x$
\answer{$y'=\displaystyle\frac{3-2x}{2y}$}

\item $xy^2-2x^3y+x^3=1$ ; Also find $\displaystyle \left.\frac{dy}{dx}\right|_{(x,y)=(1,2)}$~.
\answer{$y'=\displaystyle\frac{6x^2y-3x^2-y^2}{2xy-2x^3}$ ; $\displaystyle \left.\frac{dy}{dx}\right|_{(x,y)=(1,2)}=\frac{5}{2}$}

\item $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
\answer{$\displaystyle y'=-\frac{b^2x}{a^2y}$}

\item $\displaystyle\sin(xy)=y$
\answer{$\displaystyle y'=\frac{y\cos(xy)}{1-x\cos(xy)}$}

\item $\displaystyle \cos(x+y)+x^2=\sin y$
\answer{$\displaystyle y'=\frac{2x-\sin(x+y)}{\cos y+\sin(x+y)}$}

\item $\displaystyle \frac{\sec y}{x}=\sin x$
\answer{$\displaystyle y'=\frac{x^2\cos x+\sec y}{x\sec y\tan y}$}

\end{problemblock}

\item Consider the curve generated by the relation $\displaystyle x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$~.
\begin{enumerate}
\item Confirm that the point $\displaystyle P(-1,3\sqrt{3})$ lies on the curve.
\item Find the equation of the tangent line to the curve at the point $P$~.
\end{enumerate}
\answer{\begin{enumerate}
\item Noting that $x^{\frac{2}{3}}=\left(\sqrt[3]{x}\right)^2$ it follows that the values $x=-1$ and $y=3\sqrt{3}=(\sqrt{3})^3$ simultaneously satisfy the equation.
\item Point-slope form: $y=\sqrt{3}(x+1)+3\sqrt{3}$ , Slope-intercept form: $y=\sqrt{3}\,x+4\sqrt{3}$
\end{enumerate}}

