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

解

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

- 6 y

解答ステップ

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

x

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

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

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

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

= 0 - 6 y - 0

簡素化

= - 6 y