[SOLVED] the slope of the line tangent to the curve y^2 + (xy +1)^3=0 at (2,-1) is?

This is the best answer for the slope of the line tangent to the curve y^2 + (xy +1)^3=0 at (2,-1) is?

Differentiating the given one

2y(dy/dx) + 3{(xy+1)²}*{y + x(dy/dx)} = 0

At (2, -1) it is:

-2(dy/dx) + 3(1){-1 + 2(dy/dx)} = 0

==> -2(dy/dx) – 3 + 6(dy/dx) = 0

==> dy/dx = 3/4

This is the slope of the tangent at the point (2, -1)

I think this is the correct answer for the slope of the line tangent to the curve y^2 + (xy +1)^3=0 at (2,-1) is?

Differentiating implicitly:

2yy’ + 3(xy + 1)²*(xy’ + y) = 0

2yy’ + 3x(xy + 1)²y’ + 3y(xy + 1)² = 0

y’ = (-3y(xy + 1)²)/(2y + 3x(xy + 1)²)

So slope of tangent line at (2,-1) is:

y'(2,-1) = (3(-1)²)/(-2 + 6(1)) = 3/4

implicitly differentiate this function, and then plug the points in. Its a pretty tough function to implicitly differentiate, so you need to algebraically manipulate it a little. In the end i think you will get 3/4 as your slope.