If $x^{2} + y^3 - 2 x y^2 = 1$, find $dy/dx$ in terms of $x$ and $y$.
${dy\over dx} =$

Using your answer for $dy/dx$, fill in the following table of approximate $y$-values of points on the curve near $x=1, y=2$.

0.96	0.98	1.02	1.04

Finally, find the $y$-value for $x=0.96$ by substituting $x=0.96$ in the original equation and solving for $y$ using a computer or calculator.
$y(0.96) \approx$

How large (in magnitude) is the difference between your estimate for $y(0.96)$ using $dy/dx$ and your solution with a computer or calculator?