(\partial)/(\partial x)(x^2+2y^2-x^2y-1)

解

\frac{\partial}{\partial x} (x^{2} + 2 y^{2} - x^{2} y - 1)

2 x - 2 x y

解答ステップ

\frac{\partial}{\partial x} (x^{2} + 2 y^{2} - x^{2} y - 1)

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} (2 y^{2}) - \frac{\partial}{\partial x} (x^{2} y) - \frac{\partial}{\partial x} (1)

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

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

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

\frac{\partial}{\partial x} (1) = 0

= 2 x + 0 - 2 y x - 0

簡素化

= 2 x - 2 x y