Differentiation: A sign question.

It was a longer question but one part involved me differentiating: "-2xy."
I took this to be -2x(dy/dx) -2y, but the mark-scheme says it's "-2x(dy/dx) + 2y". Why is it +2y??

x^3 − 2xy − 4x + y^3 − 51 = 0<==== start differentiation

3x^2 − 2(x * y' + y * 1 ) − 4 + 3y^2 * y' = 0

3x^2 − 2xy' - 2y − 4 + 3y^2 * y' = 0

3x^2 - 2y − 4 = 2xy' - 3y^2 * y'

3x^2 - 2y − 4 = y' * (2x - 3y^2)

(3x^2 - 2y − 4) / (2x - 3y^2) = y' = m ====> at the point (4 , 3)

(3 * 4^2 - 2 * 3 − 4) / (2 * 4 - 3 * 3^2) = m

-2 = m <=== perpendicular as 1/2

y = mx + b <===== at (4 , 3)
3 = (1/2) * 4 + b
3 = 2 + b
3 - 2 = b
1 = b

y = (1/2)x + 1

-(1/2)x + y - 1 = 0


-x + 2y - 2 = 0

you should put the whole question down i think there may have been a negative sign the other person is incorrect .

if you differentiate -2xy

-2x*dy/dx -2y*dx/dx
= -2x(dy/dx) - 2y

now that is the answer but you may have forgotten a exponent
or there may have been a 4y from another section that adds to -2y to give 2y


without seeing the whole question I cant know for sure ..

thirdly I have many mistakes in textbooks I just found one recently in a question and asked on yahoo for comfirmation if I was correct and the book incorrect so the book may be incorrect so dont get upset if it is just argue your point see what the teacher says..

i think it must be
- [2x ( dy / dx) + 2y]