Find dy/dx for (x2y3) + 3x3y2 = 6.

To find (dy)/(dx) for the given equation (x^2y^3) + 3x^3y^2 = 6, we will use implicit differentiation, applying the chain rule and product rule where necessary. Step 1: Differentiate both sides of the equation with respect to x. (d)/(dx)((x^2y^3)) + (d)/(dx)(3x^3y^2) = (d)/(dx)(6) Step 2: Differentiate the first term, (x^2y^3). We use the chain rule: (d)/(dx)((u)) = (u) (du)/(dx), where u = x^2y^3. To find (du)/(dx) = (d)/(dx)(x^2y^3), we use the product rule: (d)/(dx)(fg) = f'g + fg'. Let f = x^2 and g = y^3. Then f' = (d)/(dx)(x^2) = 2x. And g' = (d)/(dx)(y^3) = 3y^2 (dy)/(dx) (by chain rule, since y is a function of x). So, (d)/(dx)(x^2y^3) = (2x)(y^3) + (x^2)(3y^2 (dy)/(dx)) = 2xy^3 + 3x^2y^2 (dy)/(dx). Therefore, 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. We use the product rule. Let f = 3x^3 and g = y^2. Then f' = (d)/(dx)(3x^3) = 9x^2. And g' = (d)/(dx)(y^2) = 2y (dy)/(dx). So, the derivative of the second term is: (d)/(dx)(3x^3y^2) = (9x^2)(y^2) + (3x^3)(2y (dy)/(dx)) = 9x^2y^2 + 6x^3y (dy)/(dx) Step 4: Differentiate the right side of the equation. (d)/(dx)(6) = 0 Step 5: Combine all differentiated terms into the equation. (x^2y^3)(2xy^3 + 3x^2y^2 (dy)/(dx)) + 9x^2y^2 + 6x^3y (dy)/(dx) = 0 Expand the first term: 2xy^3 (x^2y^3) + 3x^2y^2 (x^2y^3) (dy)/(dx) + 9x^2y^2 + 6x^3y (dy)/(dx) = 0 Step 6: Isolate terms containing (dy)/(dx). Move terms without (dy)/(dx) to the right side of the equation: 3x^2y^2 (x^2y^3) (dy)/(dx) + 6x^3y (dy)/(dx) = -2xy^3 (x^2y^3) - 9x^2y^2 Step 7: Factor out (dy)/(dx). (dy)/(dx) (3x^2y^2 (x^2y^3) + 6x^3y) = -2xy^3 (x^2y^3) - 9x^2y^2 Step 8: Solve for (dy)/(dx). (dy)/(dx) = (-2xy^3 (x^2y^3) - 9x^2y^2)/(3x^2y^2 (x^2y^3) + 6x^3y) Step 9: Simplify the expression by factoring out common terms from the numerator and denominator. From the numerator, factor out -xy^2: -xy^2(2y (x^2y^3) + 9x) From the denominator, factor out 3x^2y: 3x^2y(y (x^2y^3) + 2x) Substitute these back into the expression for (dy)/(dx): (dy)/(dx) = (-xy^2(2y (x^2y^3) + 9x))/(3x^2y(y (x^2y^3) + 2x)) Cancel out the common factor xy: (dy)/(dx) = (-y(2y (x^2y^3) + 9x))/(3x(y (x^2y^3) + 2x)) The final answer is (dy)/(dx) = (-y(2y (x^2y^3) + 9x))/(3x(y (x^2y^3) + 2x)). Send me the next one 📸