Find dy/dx given that y^3+y^2-5y-x^2=-4
To find dy/dx, we will differentiate both sides of the given equation with respect to x.
Using the chain rule, we have d(y^3)/dx = 3y^2 * dy/dx and d(y^2)/dx = 2y * dy/dx. Additionally, the derivative of -5y with respect to x is -5(dy/dx), and the derivative of -x^2 is -2x.
So, differentiating both sides of the equation, we have:
3y^2 * dy/dx + 2y * dy/dx - 5(dy/dx) - 2x = 0
Simplifying this equation, we can combine like terms:
dy/dx(3y^2 + 2y - 5) - 2x = 0
Now, we solve for dy/dx:
dy/dx = 2x / (3y^2 + 2y - 5)
To find dy/dx, we need to differentiate both sides of the equation with respect to x.
Differentiating the left side:
Using the chain rule, we can treat y as a function of x.
d/dx[y^3 + y^2 - 5y] = d/dx[x^2 - 4]
Differentiating each term separately:
3y^2 * dy/dx + 2y * dy/dx - 5 * dy/dx = 2x
Now, we can solve for dy/dx:
dy/dx * (3y^2 + 2y - 5) = 2x - 5y^2
dy/dx = (2x - 5y^2) / (3y^2 + 2y - 5)
So, the derivative of y with respect to x, dy/dx, is given by (2x - 5y^2) / (3y^2 + 2y - 5).