Question

sin(x+y)=y^2cosx dy/dx

Answer 1

the derivative of function y, with respect to x. sin(x+y)=y^2cosx

left side will have:
cos(x+y) * (1+y' )

the right side will have:
2y * y' * cos x + y^2(-sin x)

So we have altogether:
cos(x+y) * (1+y' )=2y * y' * cos x + y^2(-sin x)

now we just need to solve for y' .

cos(x+y)+ y' *cos(x+y) = 2y * y' * cos x + y^2(-sin x)

cos(x+y)+ y' *cos(x+y) = y' * 2ycos x + y^2(-sin x)

- y' * 2y cosx + y' *cos(x+y) = -cos(x+y) + y^2(-sin x)

y' = { y^2(-sin x) -cos(x+y) } / { cos(x +y) - 2y cos(x + y) }