(\partial)/(\partial y)(4x^3y^3+1)

解

\frac{\partial}{\partial y} (4 x^{3} y^{3} + 1)

12 x^{3} y^{2}

解答ステップ

\frac{\partial}{\partial y} (4 x^{3} y^{3} + 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} (4 x^{3} y^{3}) + \frac{\partial}{\partial y} (1)

\frac{\partial}{\partial y} (4 x^{3} y^{3}) = 12 x^{3} y^{2}

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

= 12 x^{3} y^{2} + 0

簡素化

= 12 x^{3} y^{2}