derivative y=-e^x(x^2-2x+3)

解

導関数

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

- e^{x} x^{2} - e^{x}

解答ステップ

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

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

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

積の法則を適用:

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

f = e^{x}, g = x^{2} - 2 x + 3

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

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

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

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

簡素化

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

= - e^{x} x^{2} - e^{x}