Find the slope of the tangent line to the curve

Question

(2x+4y)^1/2+(2xy)^1/2=13.4
at the point (6,5).

Answer 1

√(2x+4y) + √(2xy) = 13.4

(x+2y) / 2√(2x+4y) * (2+4y') + 1/(2√xy) * (2y + 2xy') = 0

(x+2y)/√(2x+4y) + 2(x+2y)y'/√(2x+4y) + √(y/x) + √(x/y) y' = 0

y' = -[(x+2y)/√(2x+4y)+√(y/x)]/[2/√(2x+4y) + √(x/y)]

= -(y√(x+2y)+√(xy))/(x√(x+2y)+2√(xy))

whew -- now just plug in the values at (6,5) and you have

-(20+√30)/(24+2√30)