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

解

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

6 x^{2} + 2 x y^{3}

解答ステップ

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

y

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

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

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

\frac{\partial}{\partial x} (3 y^{2}) = 0

= 6 x^{2} + 2 y^{3} x - 0

簡素化

= 6 x^{2} + 2 x y^{3}