Differential Equations - have I done this right?

Question

I usually screw up this type of question so plz check - thx.

sin(x+y) + cos(x-y) =1. What is dy/dx?

sin(x)cos(y) + cos(x)sin(y) + cos(x)cos(y) + sin(x)sin(y) = 1

cos(x)cos(y) - sin(x)sin(y)(dy/dx) - sin(x)sin(y) + cos(x)cos(y)(dy/dx) - sin(x)cos(y) - cos(x)sin(y)(dy/dx) + cos(x)sin(y) + sin(x)cos(y)(dy/dx) = 0

(dy/dx)(cos(x)cos(y) + sin(x)cos(y) - sin(x)sin(y) - cos(x)sin(y)) = sin(x)sin(y) + sin(x)cos(y) - cos(x)cos(y) - cos(x)sin(y)

dy/dx = [sin(x)sin(y) + sin(x)cos(y) - cos(x)cos(y) - cos(x)sin(y)] / [cos(x)cos(y) + sin(x)cos(y) - sin(x)sin(y) - cos(x)sin(y)]

dy/dx = [sin(x-y) - cos(x+y)] / [sin(x-y) + cos (x+y)]

Answer 1

Your work is correct, but you really don't need to expand out sin(x+y) or cos(x-y). You can use implicit differentiation from the start.

sin(x+y) + cos(x-y) = 1
cos(x+y)*(1+dy/dx) - sin(x-y)*(1-dy/dx) = 0
dy/dx*[cos(x+y)+sin(x-y)] = sin(x-y)-cos(x+y)
dy/dx = [sin(x-y)-cos(x+y)]/[cos(x+y)+sin(x-y)]

Answer 2

Your answer is correct, but you can get the answer more elegantly by using implicit differentiation.

Answer 3

Yes it's ok