(\partial)/(\partial y)(-8y^3+4xy+4x^2+9y^2)

解

\frac{\partial}{\partial y} (- 8 y^{3} + 4 x y + 4 x^{2} + 9 y^{2})

- 24 y^{2} + 18 y + 4 x

解答ステップ

\frac{\partial}{\partial y} (- 8 y^{3} + 4 x y + 4 x^{2} + 9 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} (8 y^{3}) + \frac{\partial}{\partial y} (4 x y) + \frac{\partial}{\partial y} (4 x^{2}) + \frac{\partial}{\partial y} (9 y^{2})

\frac{\partial}{\partial y} (8 y^{3}) = 24 y^{2}

\frac{\partial}{\partial y} (4 x y) = 4 x

\frac{\partial}{\partial y} (4 x^{2}) = 0

\frac{\partial}{\partial y} (9 y^{2}) = 18 y

= - 24 y^{2} + 4 x + 0 + 18 y

簡素化

= - 24 y^{2} + 18 y + 4 x