(\partial)/(\partial x)(2xyln(x))

解

\frac{\partial}{\partial x} (2 x y ln (x))

2 y (ln (x) + 1)

解答ステップ

\frac{\partial}{\partial x} (2 x y ln (x))

y

を定数として扱う

定数を除去:

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

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

積の法則を適用:

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

f = x, g = ln (x)

= 2 y (\frac{\partial}{\partial x} (x) ln (x) + \frac{\partial}{\partial x} (ln (x)) x)

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

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

= 2 y (1 \cdot ln (x) + \frac{1}{x} x)

簡素化

2 y (1 \cdot ln (x) + \frac{1}{x} x) : 2 y (ln (x) + 1)

= 2 y (ln (x) + 1)