Use implicit differentiation to find the slope of the tangent line to the curve
2(x^2) + 2xy + 2(y^3)= -18
What's the problem? If you understand implicit differentiation, it's just crank away:
4x + 2y + 2xy' + 6y^2 y' = 0
y'(2x + 6y^2) = -4x - 2y
y' = -2(2x+y)/2(x+3y^2) = -(2x+y)/(x+3y^2)
To find the slope of the tangent line to the curve, we need to use implicit differentiation.
Step 1: Differentiate both sides of the equation with respect to x. Treat y as a function of x and use the product rule and chain rule as necessary.
For the left side of the equation, differentiate each term separately:
d/dx(2(x^2)) = 4x
d/dx(2xy) = 2(dy/dx) * x + 2y
d/dx(2(y^3)) = 6(y^2) * (dy/dx)
For the right side of the equation, the derivative of a constant is zero, so it becomes:
d/dx(-18) = 0
Step 2: Combine the derivative terms and solve for dy/dx, which represents the slope of the tangent line.
Putting it all together:
4x + 2(dy/dx) * x + 2y + 6(y^2) * (dy/dx) = 0
Rearranging the equation to solve for dy/dx:
2(dy/dx) * x + 6(y^2) * (dy/dx) = -4x - 2y
Factoring out dy/dx:
(dy/dx)(2x + 6(y^2)) = -4x - 2y
Divide both sides by (2x + 6(y^2)):
dy/dx = (-4x - 2y) / (2x + 6(y^2))
This is the equation for the slope of the tangent line.