Answer is: option4
\( \frac{\sqrt{2}}{8} \)Solution:
Using the distributive property (to eliminate the parentheses) gives you \( y^2x^2 + y^3 = 3x^2 \). Now you use implicit differentiation and solve for \( y' \).
\[ 2xy^2 + x^2 \cdot 2yy' + 4y^3 y' = 6x \] \[ y'(x^2 2y + 4y^3) + 2xy^2 = 6x \] \[ y' = \frac{6x - 2xy^2}{2x^2y + 4y^3} \]
Now you need only substitute the values of \( x = 2 \) and \( y = \sqrt{2} \) into this formula to obtain:
\[ \frac{6(2) - 2(2)(\sqrt{2})^2}{2(2)^2 (\sqrt{2}) + 4(\sqrt{2})^3} = \frac{12 - 8}{8\sqrt{2} + 8\sqrt{2}} = \frac{4}{16\sqrt{2}} = \frac{1}{4\sqrt{2}} = \frac{\sqrt{2}}{8} \]