(\partial)/(\partial z)(x^2+y^2-z^2)

解

\frac{\partial}{\partial z} (x^{2} + y^{2} - z^{2})

- 2 z

解答ステップ

\frac{\partial}{\partial z} (x^{2} + y^{2} - z^{2})

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} (x^{2}) + \frac{\partial}{\partial z} (y^{2}) - \frac{\partial}{\partial z} (z^{2})

\frac{\partial}{\partial z} (x^{2}) = 0

\frac{\partial}{\partial z} (y^{2}) = 0

\frac{\partial}{\partial z} (z^{2}) = 2 z

= 0 + 0 - 2 z

簡素化

= - 2 z