(\partial)/(\partial y)(2x^2-xy-5x+1)

解

\frac{\partial}{\partial y} (2 x^{2} - x y - 5 x + 1)

- x

解答ステップ

\frac{\partial}{\partial y} (2 x^{2} - x y - 5 x + 1)

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} (x y) - \frac{\partial}{\partial y} (5 x) + \frac{\partial}{\partial y} (1)

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

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

\frac{\partial}{\partial y} (5 x) = 0

\frac{\partial}{\partial y} (1) = 0

= 0 - x - 0 + 0

簡素化

= - x