find the equation of the tangent to the curve 3xy^2 -4+2y=29 at the point (1,3)using implicit differentiation.

Question

find the equation of the tangent to the curve 3xy^2 -4+2y=29 at the point (1,3)using implicit differentiation.

Answer 1

3xy^2 -4+2y=29

3y^2 + 6xyy' + 2y' = 0
at (1,3), we have
27+18y'+2y' = 0
y' = -27/20

So, now we have a point and a slope, so the tangent line there is

y-3 = (-27/20)(x-1)