derivative of e^{2+x}-2e^x+2ex

解答

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

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

求解步骤

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

使用微分加减法定则:

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

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

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

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

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

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

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