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