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Find the slope of the tangent line to the curve 8 y^{3}+4 x^{2} y^{2}=-4 at the point (1,-1)

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Final answer:

To find the slope of the tangent line to the curve 8y^3+4x^2y^2=-4 at the point (1,-1), we can use implicit differentiation. Plugging in the coordinates of the point and solving, we find that the slope of the tangent line is 1.

Step-by-step explanation:

To find the slope of the tangent line to the curve 8y^3+4x^2y^2=-4 at the point (1,-1), we can use the concept of implicit differentiation. To do this, we differentiate both sides of the equation concerning x, treating y as a function of x. Differentiating, we get 24y^2(dy/dx) + 8x^2(2y(dy/dx)) + 8xy^2 = 0.

Next, we substitute the coordinates of the given point (1,-1) into the equation and solve for dy/dx. Plugging in x = 1 and y = -1, we have 24(-1)^2(dy/dx) + 8(1)^2(2(-1)(dy/dx)) + 8(1)(-1)^2 = 0.

Simplifying, we get 24(dy/dx) - 16(dy/dx) - 8 = 0. Combining like terms, 8(dy/dx) = 8. Dividing both sides by 8, we find that dy/dx = 1.

User Oliver Slay
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