(\partial)/(\partial y)({f}(x,y)(x+y))

解

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

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

解答ステップ

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

x

を定数として扱う

積の法則を適用:

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

f = f (x, y), g = x + y

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

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

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