Use implicit differentiation to find $y^{'}$ , and then evaluate $y^{'}$ for $x^{2} y - 3 x^{2} - 4 = 0$ at the point $(2, 4)$ .
$\begin{array}{l} y^{'} = ◻ \\ {y^{'} |}_{(2, 4)} = ◻ (Simplify your answer.) \end{array}$

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So, the final answer is: $y^{'} = 2 x (y - 3)$ and ${y^{'} |}_{(2, 4)} = 4$ .

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Step 1 ：Differentiate the given equation implicitly with respect to x. This involves applying the product rule to the term $x^{2} y$ and then differentiating the remaining terms normally. The differentiated equation is $2 x y - 6 x$ .

Step 2 ：Solve the resulting equation for $y^{'}$ to find the derivative of y with respect to x. The derivative, $y^{'}$ , is $2 x (y - 3)$ .

Step 3 ：Substitute the coordinates of the given point into the expression for $y^{'}$ to find the value of the derivative at that point. The value of $y^{'}$ at the point $(2, 4)$ is $4$ .

Step 4 ：So, the final answer is: $y^{'} = 2 x (y - 3)$ and ${y^{'} |}_{(2, 4)} = 4$ .

Use implicit differentiation to find y′, and then evaluate y′ for x2y−3x2−4=0 at the point (2,4).y′=◻y′|(2,4)=◻ (Simplify your answer. )

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Use implicit differentiation to find $y^{'}$ , and then evaluate $y^{'}$ for $x^{2} y - 3 x^{2} - 4 = 0$ at the point $(2, 4)$ .
$\begin{array}{l} y^{'} = ◻ \\ {y^{'} |}_{(2, 4)} = ◻ (Simplify your answer.) \end{array}$