(\partial)/(\partial y)((u)(Cy(2h-y)))

解

\frac{\partial}{\partial y} ((u) (C y (2 h - y)))

u C (2 h - 2 y)

解答ステップ

\frac{\partial}{\partial y} ((u) (C y (2 h - y)))

u, C, h

を定数として扱う

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= u C \frac{\partial}{\partial y} (y (2 h - y))

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y, g = 2 h - y

= u C (\frac{\partial}{\partial y} (y) (2 h - y) + \frac{\partial}{\partial y} (2 h - y) y)

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

\frac{\partial}{\partial y} (2 h - y) = - 1

= u C (1 \cdot (2 h - y) + (- 1) y)

簡素化

u C (1 \cdot (2 h - y) + (- 1) y) : u C (2 h - 2 y)

= u C (2 h - 2 y)