The equation x ^ 2 + XY + y ^ 2 = 4, why do both sides get 2x + y + XY '+ 2Y * y' = 0 by deriving x at the same time? How do you get y 'in XY' and 2Y * y '? Why do you do this? Is there any rule

The equation x ^ 2 + XY + y ^ 2 = 4, why do both sides get 2x + y + XY '+ 2Y * y' = 0 by deriving x at the same time? How do you get y 'in XY' and 2Y * y '? Why do you do this? Is there any rule

X ^ 2 + XY + y ^ 2 = 4 this is an implicit function equation, y is a function of X
So when both sides take derivatives of X at the same time, (y ^ 2) '= 2Y * y'

The equation x ^ 2 + XY + y ^ 2 = 4, why do both sides obtain 2x + y + XY '+ 2Y * y' = 0 from the derivation of X at the same time, and how do you get y 'in the last 2Y * y'?

The derivation of Y ^ 2 from X is 2Y * y '
Y ^ 2 for x = 2Y * y '(2Y for y ^ 2 first, then y, so multiply y'.)
Including the previous: XY ', y' is also derived from the derivation of X by XY
Because the derivative of y to X can not be expressed by formula, it can only be expressed by symbol y 'for the time being

If x-3y = 2x + Y-15 = 1, then x ^ 2-4xy + y ^ 2
Who is right to confirm the answer tomorrow

Substituting X - 3Y = 1, x = 3Y + 1 into 2x + Y-15 = 1, y = 2
Then x = 7
x^2-4xy+y^2 =49 - 56 + 4 = -3