Find the slope of the tangent line to the curve $-x^2 + 4xy + 4y^3 = 119$ at the point $(1, 3)$.
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The given curve is -2 + 4xy + 4y = 119. To find the derivative, we differentiate the equation with respect to x: d/dx(-2) + d/dx(4xy) + d/dx(4y) = d/dx(119) 0 + 4y + 4x(dy/dx) + 4(dy/dx) = 0 4y + 4x(dy/dx) + 4(dy/dx) = 0 4(dy/dx)(x+1) = -4y (dy/dx) = -y/(x+1) Show more…
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