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

解答

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

4 e^{2 x} + 9 e^{3 x}

求解步骤

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

使用微分加减法定则:

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

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

\frac{d}{d x} (2 e^{2 x}) = 4 e^{2 x}

\frac{d}{d x} (3 e^{3 x}) = 9 e^{3 x}

= 4 e^{2 x} + 9 e^{3 x}

x \to a lim ([f (x)]^{g (x)})

t \to 0 lim (\frac{3}{t} - \frac{3}{e ^{t} - 1})

y^{^{'}} + y = t e^{- t} + 1

\int \frac{1}{12} d x

x \to + 0 + lim (\frac{1}{x} - \frac{1}{x ^{2}})