How to Differentiate y = 3y^3 + 4x^2 + 2x + 2xy + 1 with respect to x?

To differentiate the equation y = 3y^3 + 4x^2 + 2x + 2xy + 1 with respect to x, we need to apply implicit differentiation since y is expressed in terms of x.

Let’s differentiate each term individually:

  • For the term 3y^3, using the chain rule, we get:

    d/dx(3y^3) = 9y^2(dy/dx)
  • For 4x^2, it’s straightforward:

    d/dx(4x^2) = 8x
  • For the term 2x:

    d/dx(2x) = 2
  • For the product 2xy, we need to use the product rule:

    d/dx(2xy) = 2(x(dy/dx) + y)
  • The constant 1 does not change:

    d/dx(1) = 0

Putting it all together, we have:

    dy/dx = 9y^2(dy/dx) + 8x + 2 + 2(x(dy/dx) + y)

Now, we can rearrange to solve for dy/dx:

    dy/dx - 9y^2(dy/dx) - 2x(dy/dx) = 8x + 2 + 2y

Factor out dy/dx:

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

Finally, we can solve for dy/dx:

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

This is the expression for the derivative of y with respect to x.

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