Applying sin {\displaystyle \sin } on both sides of the equation, we can implicitly differentiate x y = sin ( x + y ) {\displaystyle xy=\sin(x+y)} . Then by applying product rule and chain rule we get y + x y ′ = cos ( x + y ) ( 1 + y ′ ) ⟹ y + x y ′ = cos ( x + y ) + cos ( x + y ) y ′ ⟹ ( x − cos ( x + y ) ) y ′ = cos ( x + y ) − y ⟹ y ′ = cos ( x + y ) − y x − cos ( x + y ) . {\displaystyle {\begin{aligned}y+xy'&=\cos(x+y)(1+y')\implies \\y+xy'&=\cos(x+y)+\cos(x+y)y'\implies \\(x-\cos(x+y))y'&=\cos(x+y)-y\implies \\y'&={\frac {\cos(x+y)-y}{x-\cos(x+y)}}.\end{aligned}}}
