derivative of (x^2+xe^x)

解

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

e^{x} x^{2} + 3 e^{x} x + e^{x}

解答ステップ

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

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2} + x, g = e^{x}

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

\frac{d}{d x} (x^{2} + x) = 2 x + 1

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

= (2 x + 1) e^{x} + e^{x} (x^{2} + x)

簡素化

(2 x + 1) e^{x} + e^{x} (x^{2} + x) : e^{x} x^{2} + 3 e^{x} x + e^{x}

= e^{x} x^{2} + 3 e^{x} x + e^{x}