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

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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)

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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

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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.
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat.
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