Question Detail

Differentiate both sides of the equation implicitly with respect to x:d/dx (3x2 y) + d/dx (xy2) = d/dx (5)For d/dx (3x2 y), use the product rule: (6x)y + 3x2(dy/dx)For d/dx (xy2), use the product rule: (1)y2 + x(2y dy/dx) = y2 + 2xy(dy/dx)For d/dx (5), it's 0.So, the equation becomes:6xy + 3x2(dy/dx) + y2 + 2xy(dy/dx) = 0Group terms containing dy/dx:(3x2 + 2xy)(dy/dx) = -6xy - y2Factor out a negative sign on the right:(3x2 + 2xy)(dy/dx) = -(6xy + y2)Solve for dy/dx:dy/dx = -(y2 + 6xy) / (3x2 + 2xy).