(\partial)/(\partial x)(x^2-3xy+y^2)

解

\frac{\partial}{\partial x} (x^{2} - 3 x y + y^{2})

2 x - 3 y

解答ステップ

\frac{\partial}{\partial x} (x^{2} - 3 x y + y^{2})

y

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 x} (x^{2}) - \frac{\partial}{\partial x} (3 x y) + \frac{\partial}{\partial x} (y^{2})

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

\frac{\partial}{\partial x} (3 x y) = 3 y

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

= 2 x - 3 y + 0

簡素化

= 2 x - 3 y