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

Answer 1

xy³+y-1 = 0

Find slope, m= dy/dx, of tangent line at (1,0)

Have to use implicit derivation

y³ + 3xy²y' + y' = 0

Solve for y'

(3xy²+1)y' = -y³

y' = -y³/(3xy²+1)

at (1,0):

y' = 0/1 = 0