(\partial)/(\partial y)(8x-(x+y)^3)

解

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

- 3 x^{2} - 6 x y - 3 y^{2}

解答ステップ

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

簡素化

8 x - (x + y)^{3} : 8 x - x^{3} - 3 x^{2} y - 3 x y^{2} - y^{3}

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

\frac{\partial}{\partial y} (8 x) = 0

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

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

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

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

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

簡素化

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