derivative of (x^3-4x+1e^x)

解

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

e^{x} x^{3} + 3 e^{x} x^{2} - 4 e^{x} x - 3 e^{x}

解答ステップ

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

積の法則を適用:

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

f = x^{3} - 4 x + 1, g = e^{x}

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

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

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

= (3 x^{2} - 4) e^{x} + e^{x} (x^{3} - 4 x + 1)

簡素化

(3 x^{2} - 4) e^{x} + e^{x} (x^{3} - 4 x + 1) : e^{x} x^{3} + 3 e^{x} x^{2} - 4 e^{x} x - 3 e^{x}

= e^{x} x^{3} + 3 e^{x} x^{2} - 4 e^{x} x - 3 e^{x}