# The slope of the line tangent to the curve xy+(y+1)^2=6 at the point (2, 1) is

## the method

x dy + y dx+2(y+1)dy = 0
dy(x+2(y+1))=-ydx
dy/dx=slope= -y/(x+2(y+1))

so figure dy/dx at the given point

check my work.

## To find the slope of the line tangent to the curve at a given point, you can use the concept of implicit differentiation. The equation of the curve is given as xy + (y + 1)^2 = 6.

First, differentiate both sides of the equation with respect to x, treating y as a function of x:

d/dx [xy + (y + 1)^2] = d/dx [6]

Using the product rule and chain rule, we can find the derivatives of each term on the left side of the equation:

[d/dx (x) * y] + [x * d/dx (y)] + [d/dx [(y + 1)^2]] = 0

Now we can plug in the coordinates of the point (2, 1) into the equation to find the slope of the tangent line at that point:

[2 * 1] + [2 * dy/dx] + [(dy/dx) * 2(y + 1)] = 0

Simplifying the equation:

2 + 2(dy/dx) + 2(dy/dx)(y + 1) = 0

Now, substitute y = 1 back into the equation:

2 + 2(dy/dx) + 2(dy/dx)(1 + 1) = 0

2 + 2(dy/dx) + 4(dy/dx) = 0

6(dy/dx) = -2

(dy/dx) = -1/3

Therefore, the slope of the line tangent to the curve at the point (2, 1) is -1/3.