help find dy/dx if y=cot(x+y)

Question

working plz

Answer 1

y = cot(x+y)
y' = -csc^2(x+y) (1+y')
y' = -csc^2(x+y) - csc^2(x+y) y'
y' (1+csc^2(x+y)) = -csc^2(x+y)
y' = -csc^2(x+y)/(1+csc^2(x+y))

But, csc^2 = 1+cot^2, so

y' = -(1+y^2)/(1+1+y^2)
= -(1+y^2)/(2+y^2)

or, you can say

1/y = tan(x+y)
arctan(1/y) = x+y
1/(1 + 1/y^2))(-1/y^2)y' = 1+y'
-1/(1+y^2) y' = 1+y'
y' (1 + 1/(1+y^2)) = -1
y' (2+y^2)/(1+y^2) = -1
y' = -(1+y^2)/(2+y^2)