(\partial)/(\partial y)(x^2-xy+4y^2)

解

\frac{\partial}{\partial y} (x^{2} - x y + 4 y^{2})

8 y - x

解答ステップ

\frac{\partial}{\partial y} (x^{2} - x y + 4 y^{2})

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} (x^{2}) - \frac{\partial}{\partial y} (x y) + \frac{\partial}{\partial y} (4 y^{2})

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

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

\frac{\partial}{\partial y} (4 y^{2}) = 8 y

= 0 - x + 8 y

簡素化

= 8 y - x