Question

How do you differentiate `-2y=y^2cos(x-y)`?

Answer 1

If you mean implicitly we have that

#-2dy/dx=2y*dy/dx*cos(x-y)-y^2*sin(x-y)*[1-dy/dx]=> y^2*sin(x-y)=dy/dx*[2ycos(x-y)+2+y^2*sin(x-y)]=> dy/dx=[y^2*sin(x-y)]/[2ycos(x-y)+2+y^2*sin(x-y)]#