d/(dt)({f}(t)(t)^{-1})

解答

\frac{d}{d t} (f (t) (t)^{- 1})

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

求解步骤

\frac{d}{d t} (f (t) (t)^{- 1})

使用微分乘法定则:

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

f = f (t), g = t^{- 1}

= \frac{d}{d t} (f (t)) t^{- 1} + \frac{d}{d t} (t^{- 1}) f (t)

\frac{d}{d t} (t^{- 1}) = - \frac{1}{t ^{2}}

= \frac{d}{d t} (f (t)) t^{- 1} + (- \frac{1}{t ^{2}}) f (t)

化简

\frac{d}{d t} (f (t)) t^{- 1} + (- \frac{1}{t ^{2}}) f (t) : \frac{\frac{d}{d t} ( f ( t ) )}{t} - \frac{f ( t )}{t ^{2}}

= \frac{\frac{d}{d t} ( f ( t ) )}{t} - \frac{f ( t )}{t ^{2}}

n = 1 \sum \infty n^{n} \cdot x^{n}

y^{^{''}} + 6 y^{^{'}} + 8 = 0

\frac{d y}{d x} = 2 x (x - 4)

\frac{d y}{d x} - \frac{y}{x} = 2 x e^{x}

n = 1 \sum \infty (\frac{2 x + 1}{x})^{n}