derivative of xe^y+y^2e^x-1/3 x+4

解

\frac{d}{d x} (x e^{y} + y^{2} e^{x} - \frac{1}{3} x + 4)

e^{y} x \frac{d y}{d x} + e^{y} + e^{x} y^{2} + 2 e^{x} y \frac{d y}{d x} - \frac{1}{3}

解答ステップ

\frac{d}{d x} (x e^{y} + y^{2} e^{x} - \frac{1}{3} x + 4)

簡素化

x e^{y} + y^{2} e^{x} - \frac{1}{3} x + 4 : e^{y} x + e^{x} y^{2} - \frac{x}{3} + 4

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

和/差の法則を適用:

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

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

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

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

\frac{d}{d x} (\frac{x}{3}) = \frac{1}{3}

\frac{d}{d x} (4) = 0

= e^{y} x \frac{d y}{d x} + e^{y} + e^{x} y^{2} + 2 y \frac{d y}{d x} e^{x} - \frac{1}{3} + 0

簡素化

= e^{y} x \frac{d y}{d x} + e^{y} + e^{x} y^{2} + 2 e^{x} y \frac{d y}{d x} - \frac{1}{3}