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

解

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

6 x y

解答ステップ

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

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

\frac{\partial}{\partial y} (x^{3}) = 0

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

= 0 - 0 + 6 x y

簡素化

= 6 x y