Differentiate x^2y^3+x^3y^4=11 implicitly to find dy/dx.

2xy^3 + 3x^2y^2y' + 3x^2y^4 + 4x^3y^3y' = 0

y' = -(2xy^3 + 3x^2y^4)/(3x^2y^2 + 4x^3y^3)
= -(2y + 3xy^2)/(3x + 4x^2y)
assuming x,y not zero

Thanks a lot, it makes so much more sense now! :)

To differentiate the equation x^2y^3 + x^3y^4 = 11 implicitly, you need to use the rules of implicit differentiation. Here's how you can do it step by step:

Step 1: Differentiate both sides of the equation with respect to x.
d/dx(x^2y^3 + x^3y^4) = d/dx(11)

Step 2: Apply the product rule on each term separately.
(d/dx(x^2y^3)) + (d/dx(x^3y^4)) = 0

Step 3: Differentiate each term using the chain rule.
(2xy^3 + x^2 * 3y^2 * dy/dx) + (3x^2y^4 + x^3 * 4y^3 * dy/dx) = 0

Step 4: Simplify the equation.
2xy^3 + 3x^2y^2 * dy/dx + 3x^2y^4 + 4x^3y^3 * dy/dx = 0

Step 5: Rearrange the equation to solve for dy/dx.
Separate the terms that contain dy/dx and put all other terms on the other side of the equation:
3x^2y^2 * dy/dx + 4x^3y^3 * dy/dx = -2xy^3 - 3x^2y^4

Factor out dy/dx:
dy/dx(3x^2y^2 + 4x^3y^3) = -2xy^3 - 3x^2y^4

Now, divide both sides by (3x^2y^2 + 4x^3y^3) to solve for dy/dx:
dy/dx = (-2xy^3 - 3x^2y^4) / (3x^2y^2 + 4x^3y^3)

So, the derivative dy/dx is equal to (-2xy^3 - 3x^2y^4) / (3x^2y^2 + 4x^3y^3).