Find the slope of the tangent line for the curve xy^3-yx^3=6 at the point(1,1)

1 answer

To find the slope of the tangent line at a given point on a curve, we need to find the derivative of the curve with respect to x and evaluate it at the given point.

First, rearrange the equation to isolate y:
xy^3 - yx^3 = 6
y(x - x^3) = 6
y = 6 / (x - x^3)

Now, take the derivative of y with respect to x:
dy/dx = d(6 / (x - x^3))/dx

Using the quotient rule,
dy/dx = (0 - 6 * (1 - 3x^2)) / (x - x^3)^2
= -6(1 - 3x^2) / (x - x^3)^2

To find the slope at the point (1,1), substitute x = 1 into the expression for the derivative:
dy/dx = -6(1 - 3(1)^2) / (1 - 1^3)^2
= -6(1 - 3)/ (1 - 1)^2
= -6(-2)/0

The expression is undefined when the denominator is zero, indicating that the slope at the point (1,1) is undefined. This means that the tangent line at that point is vertical.