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

解

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

3 y^{2} - 6 x y

解答ステップ

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

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

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

\frac{\partial}{\partial y} (3 x y^{2}) = 6 x y

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

= 0 - 6 x y + 3 y^{2}

簡素化

= 3 y^{2} - 6 x y