For the equation given below,evaluate y' at the point (1,2) 4x^3y-2x^2=6 y'at(1,2)= square

To evaluate \( y' \) at the point \((1, 2)\) for the given equation \( 4x^3 y - 2x^2 = 6 \), we need to implicitly differentiate the equation with respect to \( x \).<br /><br />Starting with the given equation:<br />\[ 4x^3 y - 2x^2 = 6 \]<br /><br />Differentiate both sides with respect to \( x \):<br />\[ \frac{d}{dx}(4x^3 y) - \frac{d}{dx}(2x^2) = \frac{d}{dx}(6) \]<br /><br />Using the product rule on \( 4x^3 y \):<br />\[ 4 \left( 3x^2 y + x^3 y' \right) - 4x = 0 \]<br /><br />Simplify:<br />\[ 12x^2 y + 4x^3 y' - 4x = 0 \]<br /><br />Solve for \( y' \):<br />\[ 4x^3 y' = -12x^2 y + 4x \]<br />\[ y' = \frac{-12x^2 y + 4x}{4x^3} \]<br />\[ y' = \frac{-3x y + x}{x^2} \]<br />\[ y' = \frac{-3y + 1}{x} \]<br /><br />Now, substitute \( x = 1 \) and \( y = 2 \) into the expression for \( y' \):<br />\[ y' \bigg|_{(1, 2)} = \frac{-3(2) + 1}{1} \]<br />\[ y' \bigg|_{(1, 2)} = \frac{-6 + 1}{1} \]<br />\[ y' \bigg|_{(1, 2)} = \frac{-5}{1} \]<br />\[ y' \bigg|_{(1, 2)} = -5 \]<br /><br />Thus, the value of \( y' \) at the point \((1, 2)\) is:<br />\[ \boxed{-5} \]