(\partial)/(\partial x)(x^4+3xy^2-3y^2)

解

\frac{\partial}{\partial x} (x^{4} + 3 x y^{2} - 3 y^{2})

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

解答ステップ

\frac{\partial}{\partial x} (x^{4} + 3 x y^{2} - 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} (x^{4}) + \frac{\partial}{\partial x} (3 x y^{2}) - \frac{\partial}{\partial x} (3 y^{2})

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

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

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

= 4 x^{3} + 3 y^{2} - 0

簡素化

= 4 x^{3} + 3 y^{2}