Use implicit differentiation to find the slope of the tangent line to the curve –2x^2–2xy–1y^3=–3 at the point (–1–1).
To find the slope of the tangent line to the given curve at the point (-1, -1) using implicit differentiation, we need to follow these steps:
Step 1: Differentiate both sides of the equation with respect to x. Treat y as a function of x and apply the product and chain rule as needed.
For the equation -2x^2 - 2xy - 1y^3 = -3, the derivative with respect to x would be:
d/dx(-2x^2) - d/dx(2xy) - d/dx(y^3) = d/dx(-3)
Simplifying each term using the chain rule and product rule:
-4x - 2y - 2x(dy/dx) - 2y - 3y^2(dy/dx) = 0
Step 2: Substitute the x and y values of the given point (-1, -1) into the differentiated equation.
Plugging in x = -1 and y = -1 into the equation:
-4(-1) - 2(-1) - 2(-1)(dy/dx) - 2(-1) - 3(-1)^2(dy/dx) = 0
4 + 2 - 2(dy/dx) + 2 + 3(dy/dx) = 0
Simplifying the equation further:
8 + dy/dx = 0
Step 3: Solve for dy/dx, which represents the slope of the tangent line.
Isolating dy/dx, we subtract 8 from both sides:
dy/dx = -8
The slope (dy/dx) of the tangent line to the curve -2x^2 - 2xy - 1y^3 = -3 at the point (-1, -1) is -8.