Find dy/dx for the equation sin(x2y3) + 3x3y2 = 6 using implicit differentiation.

Step 1: Differentiate both sides of the equation with respect to x. The given equation is (x^2y^3) + 3x^3y^2 = 6. (d)/(dx)[(x^2y^3) + 3x^3y^2] = (d)/(dx)[6] (d)/(dx)[(x^2y^3)] + (d)/(dx)[3x^3y^2] = 0 Step 2: Differentiate the first term, (x^2y^3), using the chain rule. Let u = x^2y^3. Then (d)/(dx)[(u)] = (u) (du)/(dx). To find (du)/(dx) = (d)/(dx)[x^2y^3], we use the product rule, where f = x^2 and g = y^3. (d)/(dx)[x^2y^3] = (d)/(dx)[x^2] · y^3 + x^2 · (d)/(dx)[y^3] = (2x)y^3 + x^2(3y^2 (dy)/(dx)) = 2xy^3 + 3x^2y^2 (dy)/(dx) So, the derivative of the first term is: (d)/(dx)[(x^2y^3)] = (x^2y^3)(2xy^3 + 3x^2y^2 (dy)/(dx)) Step 3: Differentiate the second term, 3x^3y^2, using the product rule. Here, f = 3x^3 and g = y^2. (d)/(dx)[3x^3y^2] = (d)/(dx)[3x^3] · y^2 + 3x^3 · (d)/(dx)[y^2] = (9x^2)y^2 + 3x^3(2y (dy)/(dx)) = 9x^2y^2 + 6x^3y (dy)/(dx) Step 4: Substitute these derivatives back into the equation from Step 1. (x^2y^3)(2xy^3 + 3x^2y^2 (dy)/(dx)) + 9x^2y^2 + 6x^3y (dy)/(dx) = 0 Step 5: Expand the equation and group terms containing (dy)/(dx). 2xy^3 (x^2y^3) + 3x^2y^2 (x^2y^3) (dy)/(dx) + 9x^2y^2 + 6x^3y (dy)/(dx) = 0 Move terms without (dy)/(dx) to the right side: 3x^2y^2 (x^2y^3) (dy)/(dx) + 6x^3y (dy)/(dx) = -2xy^3 (x^2y^3) - 9x^2y^2 Factor out (dy)/(dx) from the left side: (3x^2y^2 (x^2y^3) + 6x^3y) (dy)/(dx) = -2xy^3 (x^2y^3) - 9x^2y^2 Step 6: Solve for (dy)/(dx). (dy)/(dx) = (-2xy^3 (x^2y^3) - 9x^2y^2)/(3x^2y^2 (x^2y^3) + 6x^3y) Step 7: Simplify the expression by factoring common terms from the numerator and denominator. Factor -xy^2 from the numerator: Numerator = -xy^2(2y (x^2y^3) + 9x) Factor 3xy from the denominator: Denominator = 3xy(xy (x^2y^3) + 2x^2) Now, substitute these factored forms back into the expression for (dy)/(dx): (dy)/(dx) = (-xy^2(2y (x^2y^3) + 9x))/(3xy(xy (x^2y^3) + 2x^2)) Cancel xy from the numerator and denominator (assuming $