Find dy/dx by implicit differentation?

Question

1.square root of (xy)=x-2y

2. 2sinxcosy=1

Answer 1

1. (xy)^1/2 = x - 2y
[(1/2)(xy)^(-1/2)](xdy + y) = 1 - 2dy
[(1/2)(xy)^(-1/2)](xdy) + [1/2)(xy)^(-1/2)](y) = 1 - 2dy
[1/2)(xy)^(-1/2)](xdy) + 2dy = 1 - [(1/2)(xy)^(-1/2)](y)
dy [(x/2)(xy)^(-1/2) + 2] = 1 - [(y/2)(xy)^(-1/2)]
dy = {1 - [(y/2)(xy)^(-1/2)]} / [(x/2)(xy)^(-1/2) + 2]

2. 2sinxcosy = 1
2sinx(-siny)dy + cosycosx = 0
cosycosx = 2sinxsinydy
dy = cosycosx/2sinxsiny
dy = cotxcoty/2