Find the slope of the tangent line to the curve
(3x+3y−27)^(1/2)+(2xy−39)^(1/2)=8
at the point (8,4).
√(3x+3y-27) + √(2xy-39) = 8
(3+3y')/2√(3x+3y-27) + (2y + 2xy')/2√(2xy-39) = 0
At (8,4)
(3+3y')/2√9 + (8+16y')/2√25 = 0
(1+y')/2 + (4+8y')/5 = 0
5(1+y') + 2(4+8y') = 0
5 + 5y' + 8 + 16y' = 0
21y' = -13
y' = -13/21