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

解

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

2 x^{2} y - x

解答ステップ

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

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

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

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

= 2 x^{2} y - x + 0

簡素化

= 2 x^{2} y - x