d/(dt)((x)(2te^2+e^t))

解答

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

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

求解步骤

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

将

x

视为常数

将常数提出:

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

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

使用微分加减法定则:

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

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

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

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

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

e x p an d \frac{( - 1 ) ^{n + 2} ( n + 1 ) ( n + 2 )}{2}

e x p an d \frac{x ( x ^{2} - 2 )}{y ^{2}}

e x p an d \frac{y ( x - π )}{x ( 1 - p ^{2} )}

e x p an d \frac{( 6 + b ) ^{2}}{2 t - 6}

e x p an d so l v e (lo g_{10} (N) = 0.3 + 2 t, N)