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

解

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

2 x - 3 y^{2}

解答ステップ

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

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

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

\frac{\partial}{\partial y} (y^{3}) = 3 y^{2}

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

= 0 - 3 y^{2} + 2 x

簡素化

= 2 x - 3 y^{2}