d/(dt)((t^2+2)/(t^4-3t^2+1))

解答

\frac{d}{d t} (\frac{t ^{2} + 2}{t ^{4} - 3 t ^{2} + 1})

\frac{- 2 t ^{5} - 8 t ^{3} + 14 t}{( t ^{4} - 3 t ^{2} + 1 ) ^{2}}

求解步骤

\frac{d}{d t} (\frac{t ^{2} + 2}{t ^{4} - 3 t ^{2} + 1})

使用微分除法定则:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d t} ( t ^{2} + 2 ) ( t ^{4} - 3 t ^{2} + 1 ) - \frac{d}{d t} ( t ^{4} - 3 t ^{2} + 1 ) ( t ^{2} + 2 )}{( t ^{4} - 3 t ^{2} + 1 ) ^{2}}

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

\frac{d}{d t} (t^{4} - 3 t^{2} + 1) = 4 t^{3} - 6 t

= \frac{2 t ( t ^{4} - 3 t ^{2} + 1 ) - ( 4 t ^{3} - 6 t ) ( t ^{2} + 2 )}{( t ^{4} - 3 t ^{2} + 1 ) ^{2}}

乘开

2 t (t^{4} - 3 t^{2} + 1) - (4 t^{3} - 6 t) (t^{2} + 2) : - 2 t^{5} - 8 t^{3} + 14 t

= \frac{- 2 t ^{5} - 8 t ^{3} + 14 t}{( t ^{4} - 3 t ^{2} + 1 ) ^{2}}

\frac{d}{d x} (4 sin^{2} (2 x))

\frac{\partial}{\partial y} (4 x^{3} y^{3} - 4)

y^{^{''}} - 2 y^{^{'}} - 4 y = 0

x \to 2 lim (6 x^{3} + 1)

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