derivative of 4x-x^2y+y^3

解

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

4 - 2 x y - x^{2} \frac{d y}{d x} + 3 y^{2} \frac{d y}{d x}

解答ステップ

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

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (4 x) = 4

\frac{d}{d x} (x^{2} y) = 2 x y + \frac{d y}{d x} x^{2}

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

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

簡素化

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

= 4 - 2 x y - x^{2} \frac{d y}{d x} + 3 y^{2} \frac{d y}{d x}