d/(dt)((x)(xt+x)e^t)

解答

\frac{d}{d t} ((x) (x t + x) e^{t})

x (2 e^{t} x + e^{t} x t)

求解步骤

\frac{d}{d t} ((x) (x t + x) e^{t})

将

x

视为常数

将常数提出:

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

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

使用微分乘法定则:

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

f = x t + x, g = e^{t}

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

\frac{d}{d t} (x t + x) = x

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

= x (x e^{t} + e^{t} (x t + x))

乘开

x e^{t} + e^{t} (x t + x) : 2 e^{t} x + e^{t} x t

= x (2 e^{t} x + e^{t} x t)

e x p an d (e^{x} - e^{- 5 x})^{2}

e x p an d \frac{1}{x ( x + b )}

e x p an d e^{2 x} \cdot 2 (e^{x} + e^{- x})

e x p an d \frac{( - 2 x + 4 ) ^{7}}{( 3 x - 5 ) ^{5}}

e x p an d x d y - 4 (y + x^{6}) d x