derivative of e^{x^4}+7x

解答

\frac{d}{d x} (e^{x^{4}} + 7 x)

e^{x^{4}} \cdot 4 x^{3} + 7

求解步骤

\frac{d}{d x} (e^{x^{4}} + 7 x)

使用微分加减法定则:

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

= \frac{d}{d x} (e^{x^{4}}) + \frac{d}{d x} (7 x)

\frac{d}{d x} (e^{x^{4}}) = e^{x^{4}} \cdot 4 x^{3}

\frac{d}{d x} (7 x) = 7

= e^{x^{4}} \cdot 4 x^{3} + 7

n or ma l f (x) = x^{2} + 16, at x = 3

\int \frac{e ^{8 x}}{e ^{8 x} + 6} d x

y \to 1 lim (\frac{2 y ^{2} + 3}{y - 1})

\frac{d}{d t} (2 e^{- t} t - e^{- t} t^{2})

\frac{d}{d x} ((9 x^{6} + 4 x^{3})^{4})