How to differentiate y = 3y³ + 4x² + 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 = 3y³ + 4x² + 2x + 2xy + 1 with respect to x is dy / dx = (8x + 2y + 2) / (1 - 2x - 9y²).

Let's understand the solution in detail.

Explanation:

We use the methods of implicit differentiation to differentiate the function y = 3y³ + 4x² + 2x + 2xy + 1.

Now, differentiating both sides with respect to x:

⇒ dy / dx = d [3y³ + 4x² + 2x + 2xy + 1] / dx 

⇒ dy / dx = 9y² dy/dx + 8x + 2 + d/dx (2xy) + 0 (Using chain rule of differentiation)

Using the product rule on the term 2xy:

⇒ dy / dx = (9y² . dy/dx) + 8x + 2 + (2x . dy/dx) + 2y

⇒ dy / dx = (9y² . dy/dx) + (2x . dy/dx) + 8x + 2 + 2y

⇒ dy / dx - (9y² . dy/dx) - (2x . dy/dx) = 8x + 2 + 2y

⇒ dy / dx [1 - 9y² - 2x] = 8x + 2y + 2

⇒ dy / dx = (8x + 2y + 2) / (1 - 2x - 9y²)

Hence, the derivative of y = 3y³ + 4x² + 2x + 2xy + 1 with respect to x is dy / dx = (8x + 2y + 2) / (1 - 2x - 9y²)