Question

how would i go about finding the tangent line to the curve of sqrt(1+4x^2y^2)-3xy=-18 with the points (3,6)?

Answer 1

find slope at (3,6)
show line with that slope though (3,6)

√(1+4x^2y^2)-3xy = -18
1/2√(1+4x^2y^2) * (8xy^2 + 8x^2y y') - 2y - 2xy' = 0
8xy^2/2√(1+4x^2y^2) - 2y = y'(8x^2y/2√(1+4x^2y^2) - 2x) = 0

y' = (2y - 8xy^2/2√(1+4x^2y^2) )/(8x^2y/2√(1+4x^2y^2) - 2x)

plug in (3,6) for (x,y) and evaluate for slope.

find line.