(\partial)/(\partial x)(x^2y+2y^2-6x+3y-4)

解

\frac{\partial}{\partial x} (x^{2} y + 2 y^{2} - 6 x + 3 y - 4)

2 x y - 6

解答ステップ

\frac{\partial}{\partial x} (x^{2} y + 2 y^{2} - 6 x + 3 y - 4)

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} y) + \frac{\partial}{\partial x} (2 y^{2}) - \frac{\partial}{\partial x} (6 x) + \frac{\partial}{\partial x} (3 y) - \frac{\partial}{\partial x} (4)

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

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

\frac{\partial}{\partial x} (6 x) = 6

\frac{\partial}{\partial x} (3 y) = 0

\frac{\partial}{\partial x} (4) = 0

= 2 y x + 0 - 6 + 0 - 0

簡素化

= 2 x y - 6