if -xy^3-1-y=0 then find the equations of all tangent lines to the curve when y=1. implicit derivative at a point

Answer 1

Just for convenience let me rewrite the equation as

xy³ + 1 + y = 0.

First we find the points where y = 1, i.e., which values of x satisfy

the equation with y=1

x + 1 + 1 = 0

x = -2.

Thus the curve has only one point where y=1, which is (-2, 1).

The tangent will be of the form

y = px + q, where

p = dy/dx and

q will be the result of constraining the line to go though the point (-2, 1).

To find dy/dx by implicit differentiation, we differentiate the equation by dx,

while treating y as a function of x:

d/dx(xy³ + 1 + y) = d/dx(0)

y³ + 3xy²dy/dx + dy/dx = 0

dy/dx = -y³/(3xy² + 1)

At the point x = -2, y = 1

dy/dx = 1/5

The tangent will be of the form

y = x/5 + q

At the point (-2, 1)

1 = -2/5 + q

q = 7/5

There is just one tangent line

y = x/5 + 7/5