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\chead{\color{black} \LARGE Implicit Differentiation}
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\lfoot{\small \copyright \; Hidegkuti, 2015}
\rfoot{\small Last revised: November 3, 2015}
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\begin{document}


\begin{enumerate}
\item In case of each of the relations given, express $y^{\prime }$ in terms
of $x$ and $y.$ 
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a) \ $x^{3}+y^{3}=30$

b) \ $x^{2}-y^{2}=xy$

c) \ $x^{4}+y^{4}=x^{2}y^{2}$

d) \ $\sqrt{x^{2}+y^{2}}=y^{3}-8$ $\allowbreak $

e) \ $\ln \left( x+y\right) =x^{2}+y^{2}$ 

f) \ $x^{2}-y^{2}=\tan x$

g) \ $x^{2}-y^{2}=\tan y$

h) \ $\sin x-\cos y=\ln x+\ln y$

i) \ $e^{\sin x}=\dfrac{x}{y}$ \ 
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\item Find the equation of the tangent line drawn to the graph of $%
3x^{2}+4x+2y^{2}-31y+16xy+157=0$ \ at the point $\left( -1,4\right) $.

\item Consider the relation determined by the equation $6y+xy\left(
y-x-5\right) =6x+30$. \ Find an equation for all tangent line(s) drawn to
the graph of the relation at \ $x=2$.

\item Bernoulli's Lemniscate, shown on the picture below is determined by
the equation\ \newline
$\left( x^{2}+y^{2}\right) ^{2}=18\left( x^{2}-y^{2}\right) $. \ Compute the
equation of the tangent line drawn to the lemniscate at the point $\left( 2,%
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\end{enumerate}

\begin{center}
{\Large Answers}
\end{center}

\bigskip 

\begin{enumerate}
\item a) \ $y^{\prime }=-\dfrac{x^{2}}{y^{2}}$ \ \ \ \ \ \ \ b) \ $y^{\prime
}=\dfrac{-2x+y}{-x-2y}$ \ \ \ \ \ \ \ \ c) \ $\dfrac{-2x^{3}+xy^{2}}{%
2y^{3}-x^{2}y}$\ \ \ \ \ \ \ \ d) \ $y^{\prime }=\dfrac{1-2x^{2}-2xy}{%
2xy+2y^{2}-1}$ \ \ \ \ \ \ \ \ e) \ $y^{\prime }=\dfrac{x}{3y^{2}\sqrt{%
x^{2}+y^{2}}-y}$\ \ \ \ \ \ \ \ \ f) \ $y^{\prime }=\dfrac{2x-\sec ^{2}x}{2y}
$ \ \ \ \ \ \ \ \ \ g) \ $\dfrac{2x}{2y+\sec ^{2}y}$ \ \ \ \ \ \ \ \ h) \ $%
y^{\prime }=\dfrac{-xy\cos x+y}{xy\sin y-x}$ \ \ \ \ \ \ \ \ i) \ $y^{\prime
}=\dfrac{y}{x}-\dfrac{y^{2}}{x}\left( \cos x\right) e^{\sin x}$

\item $y=2x+6$

\item $y=x+5$ \ \ \ and \ \ \ \ $y=\dfrac{3}{2}x-6$

\item $\dfrac{\sqrt{2}}{5}\left( x-2\right) =y-\sqrt{2}$
\end{enumerate}

\end{document}