(\partial)/(\partial y)(xe^{xy}-2y+1)

解

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

e^{x y} x^{2} - 2

解答ステップ

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

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

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

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

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

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

簡素化

= e^{x y} x^{2} - 2