(\partial)/(\partial y)(8xy+8y^2-2x+2y-1)

解

\frac{\partial}{\partial y} (8 x y + 8 y^{2} - 2 x + 2 y - 1)

8 x + 16 y + 2

解答ステップ

\frac{\partial}{\partial y} (8 x y + 8 y^{2} - 2 x + 2 y - 1)

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 y) + \frac{\partial}{\partial y} (8 y^{2}) - \frac{\partial}{\partial y} (2 x) + \frac{\partial}{\partial y} (2 y) - \frac{\partial}{\partial y} (1)

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

\frac{\partial}{\partial y} (8 y^{2}) = 16 y

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

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

\frac{\partial}{\partial y} (1) = 0

= 8 x + 16 y - 0 + 2 - 0

簡素化

= 8 x + 16 y + 2