Derivative and graph mismatch Using the implicit function $(x^2+y^2-1)^3=x^2y^3$ it can be shown that $y'=\frac{2xy^3-6x(x^2+y^2-1)^2}{6y(x^2+y^2-1)^2-3x^2y^2}$ but when I evaluate it for the point (1,0) I get $y'(1,0)=\frac{0}{0}$ even though the slope of the tangent line is 2 at that point.
Any ideas?
Garth
 A: Assuming you did all the algebra correctly, I haven't checked: You made an error in the algebra! Heh.
When you differentiated you got $$Q(x,y)y'+P(x,y)=0,\quad(*)$$where $P$ and $Q$ are two nasty polynomials. Then you converted that to $$y'=-\frac{P(x,y)}{Q(x,y)}.\quad(**)$$
But going from (*) to ($**$) is only valid if $Q(x,y)\ne0$. So you haven't actually shown that (**) holds, you've shown that $(**)$ holds at points where $Q\ne0$. Again assuming that you did everything else right, at $(1,0)$ equation (*) becomes $$0(2)=0,$$which is no problem since $0$ times $2$ does equal $0$. But $0(2)=0$ does not imply $2=0/0$.
A: Instead $(x^2+y^2-1)^3=x^2y^3$ we can take $x^2+y^2-1=x^{2/3}y$ near of $(1,0)$, hence 
\begin{align*}
\left(y-\frac{1}{2}x^{2/3}\right)^2&=1-x^2+\frac{x^{4/3}}{4}\\
y&=-\left(1-x^2+\frac{x^{4/3}}{4}\right)^{1/2}+\frac{x^{2/3}}{2}\\
y'&=-\frac{1}{2}\left(1-x^2+\frac{x^{4/3}}{4}\right)^{-1/2}\left(-2x+\frac{x^{1/3}}{3}\right)+\frac{1}{3}x^{-1/3}
\end{align*}
Then
\begin{align*}
y'(1)&=-\frac{1}{2}\left(1-1+\frac{1}{4}\right)^{-1/2}\left(-2+\frac{1}{3}\right)+\frac{1}{3}\\
&=-\frac{1}{2}(2)\left(-\frac{5}{3}\right)+\frac{1}{3}\\
&=2
\end{align*}
