(\partial)/(\partial x)(2y^2-3x^2-z)

解

\frac{\partial}{\partial x} (2 y^{2} - 3 x^{2} - z)

- 6 x

解答ステップ

\frac{\partial}{\partial x} (2 y^{2} - 3 x^{2} - z)

y, z

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 x} (2 y^{2}) - \frac{\partial}{\partial x} (3 x^{2}) - \frac{\partial}{\partial x} (z)

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

\frac{\partial}{\partial x} (3 x^{2}) = 6 x

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

= 0 - 6 x - 0

簡素化

= - 6 x