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

解

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

3 x^{2} - 4 y

解答ステップ

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

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

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

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

= 0 + 3 x^{2} - 4 y

簡素化

= 3 x^{2} - 4 y