Question

What is the value of dy/dx for the equation y^4 + 4 = 3x^2 - 3y^2 , given that the derivative dy/ dx exists and is equal to 0? According to the implication, if dy/dx =0, then the tangent line at that point is vertical.

Answer 1

The value of dy/dx for the equation y^4 + 4 = 3x^2 - 3y^2, given that dy/dx exists and is equal to 0, is 0.

Step-by-step explanation:

To find the value of dy/dx for the equation y^4 + 4 = 3x^2 - 3y^2, given that dy/dx exists and is equal to 0, we need to take the derivative of both sides with respect to x. Let's start by differentiating both sides of the equation:

d(y^4 + 4)/dx = d(3x^2 - 3y^2)/dx

Using the power rule and chain rule, we can simplify the derivatives:

4y^3(dy/dx) = 6x - 6y(dy/dx)

Since we are given that dy/dx = 0, we can substitute this value into the equation:

4y^3(0) = 6x - 6y(0)

0 = 6x

Therefore, the value of dy/dx for the given equation when dy/dx = 0 is 0.