Find the tangent line to the curve xy3+2y−2x=1 through the point (x,y)=(1,1).

(Enter ∗ for multiplication: type 2*x for 2x. Enter / for division: type 1/2 for 12.)

Tangent Line y=

xy^3 + 2y - 2x = 1

y^3 + 3xy^2 y' + 2y' - 2 = 0
y' = (2-y^3)/(3xy^2+2)
At (1,1), y' = 2/5

Now you have a point: (1,1) and a slope: 2/5

I'm sure you know the point-slope form of a line.

y´=1/5*x-4/5

To find the equation of the tangent line to a curve at a given point, we need to find the slope of the tangent line and the coordinates of the given point.

First, let's find the slope of the tangent line at the point (1, 1). The slope of a tangent line is equal to the derivative of the function with respect to x evaluated at that point.

The given equation is xy^3 + 2y - 2x = 1. Let's rearrange this equation to express y in terms of x: xy^3 + 2y = 2x + 1.

Differentiating both sides with respect to x, we get:
d/dx(xy^3 + 2y) = d/dx(2x + 1)
y^3 + 3xy^2 * dy/dx + 2dy/dx = 2
(3xy^2 + 2) * dy/dx = 2 - y^3
dy/dx = (2 - y^3) / (3xy^2 + 2)

Now, substitute the given point (x, y) = (1, 1) into the equation to find the slope at that point:
dy/dx = (2 - 1^3) / (3(1)(1)^2 + 2)
dy/dx = 1 / 5

So the slope of the tangent line at the point (1, 1) is 1/5.

Now, we have the slope of the tangent line and a point on the line (1, 1). We can use the point-slope form of a linear equation to find the equation of the tangent line.

The point-slope form is: y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point.

Substituting the values, we get:
y - 1 = (1/5)(x - 1)

Multiplying through by 5 to clear the fraction:
5(y - 1) = x - 1

Expanding:
5y - 5 = x - 1

Simplifying, we get:
x - 5y = -4

Therefore, the equation of the tangent line to the curve xy^3 + 2y - 2x = 1 through the point (1, 1) is:
x - 5y = -4