(\partial)/(\partial y)(x^2y-2x^2-4y^2)

解

\frac{\partial}{\partial y} (x^{2} y - 2 x^{2} - 4 y^{2})

x^{2} - 8 y

解答ステップ

\frac{\partial}{\partial y} (x^{2} y - 2 x^{2} - 4 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) - \frac{\partial}{\partial y} (2 x^{2}) - \frac{\partial}{\partial y} (4 y^{2})

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

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

\frac{\partial}{\partial y} (4 y^{2}) = 8 y

= x^{2} - 0 - 8 y

簡素化

= x^{2} - 8 y