3xy^3+4xy=63
3y^3 + 9xy^2 y' + 4y + 4xy' = 0
y'(9xy^2+4x) = -(3y^3+4y)
y' = -(3y^3+4y)/(9xy^2+4x)
So, at (9,1) y' = -7/117
Now you have a point and a slope, so the line is
y-1 = -7/117 (x-9)
Refer to
http://www.wolframalpha.com/input/?i=plot+3xy%5E3%2B4xy%3D63,+y-1+%3D+-7%2F117+(x-9)+for+0%3C%3Dx%3C%3D20
Use implicit differentiation to find an equation of the tangent line to the curve 3xy^3+4xy=63 at the point (9,1)(9,1).
1 answer
1 answer