Final answer:
To find the slope of the tangent line to the given curve at the point (1, -1), we use implicit differentiation, resulting in a slope of -10/7.
Step-by-step explanation:
To find the slope of the tangent line to the curve given by the equation 1y3 + 5x2y2 = 4 at the point (1, -1), we must first differentiate the equation with respect to x using implicit differentiation. This involves differentiating both sides of the equation, while remembering to apply the chain rule to the y terms as they are functions of x.
When differentiating, we get:
3y2 dy/dx + 5(2xy2 + x2·2yy' ) = 03y2 dy/dx + 10xy2 + 10x2y dy/dx = 0dy/dx(3y2 + 10x2y) = -10xy2dy/dx = -10xy2 / (3y2 + 10x2y)
Plugging in the coordinates of the given point (1, -1):
dy/dx|(1, -1) = -10(1)(-1)2 / (3(-1)2 + 10(1)2(-1))dy/dx|(1, -1) = 10 / (3 - 10)dy/dx|(1, -1) = -10/7
Therefore, the slope of the tangent line at the point (1, -1) is -10/7.