# How to differentiate y = 3y^{3} + 4x^{2} + 2x + 2xy + 1 with respect to x?

Differentiation is one of the most important concepts in calculus. The slopes of different curves at different points can be found using differentiation.

### Answer: The derivative of y = 3y^{3} + 4x^{2} + 2x + 2xy + 1 with respect to x is dy / dx = (8x + 2y + 2) / (1 - 2x - 9y^{2}).

Let's understand the solution in detail.

**Explanation:**

We use the methods of implicit differentiation to differentiate the function y = 3y^{3} + 4x^{2} + 2x + 2xy + 1.

Now, differentiating both sides with respect to x:

⇒ dy / dx = d / dx [3y^{3} + 4x^{2} + 2x + 2xy + 1]

⇒ dy / dx = 9y^{2} dy/dx + 8x + 2 + d/dx (2xy) + 0

Using the product rule on the term 2xy:

⇒ dy / dx = (9y^{2} . dy/dx) + 8x + 2 + (2x . dy/dx) + 2y

⇒ dy / dx = (9y^{2} . dy/dx) + 8x + 2 + (2x . dy/dx) + 2y

⇒ dy / dx - (9y^{2} . dy/dx) - (2x . dy/dx) = 8x + 2 + 2y

⇒ dy / dx [1 - 9y^{2} - 2x] = 8x + 2y + 2

⇒ dy / dx = (8x + 2y + 2) / (1 - 2x - 9y^{2})