(\partial)/(\partial x)((u(x,y))(x+y))

解

\frac{\partial}{\partial x} ((u (x, y)) (x + y))

u ((, y \cdot 1) (x + y) + 1 \cdot (x, y))

解答ステップ

\frac{\partial}{\partial x} ((u (x, y)) (x + y))

u, y

を定数として扱う

定数を除去:

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

= u \frac{\partial}{\partial x} ((x, y) (x + y))

積の法則を適用:

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

f = x, y, g = x + y

= u (\frac{\partial}{\partial x} (x, y) (x + y) + \frac{\partial}{\partial x} (x + y) (x, y))

\frac{\partial}{\partial x} (x, y) = (, y \cdot 1)

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

= u ((, y \cdot 1) (x + y) + 1 \cdot (x, y))