(\partial)/(\partial x)(f(x,y)(xy))

解

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

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

解答ステップ

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

y

を定数として扱う

定数を除去:

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

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

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

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

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