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

解

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

- x^{2} - x + 2 y

解答ステップ

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

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

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

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

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

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

簡素化

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