(\partial)/(\partial y)(x^2-2xy+y^2)

解

\frac{\partial}{\partial y} (x^{2} - 2 x y + y^{2})

2 y - 2 x

解答ステップ

\frac{\partial}{\partial y} (x^{2} - 2 x y + y^{2})

x

を定数として扱う

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial y} (x^{2}) - \frac{\partial}{\partial y} (2 x y) + \frac{\partial}{\partial y} (y^{2})

\frac{\partial}{\partial y} (x^{2}) = 0

\frac{\partial}{\partial y} (2 x y) = 2 x

\frac{\partial}{\partial y} (y^{2}) = 2 y

= 0 - 2 x + 2 y

簡素化

= 2 y - 2 x