(\partial)/(\partial z)(ln(1+xy)-z)

解

\frac{\partial}{\partial z} (ln (1 + x y) - z)

- 1

解答ステップ

\frac{\partial}{\partial z} (ln (1 + x y) - z)

x, y

を定数として扱う

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial z} (ln (1 + x y)) = 0

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

= 0 - 1

簡素化

= - 1