derivative of 1/((x^2-1(x^2+3x+3)))

解答

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

- \frac{4 x ^{3} + 9 x ^{2} + 4 x - 3}{( x ^{2} - 1 ) ^{2} ( x ^{2} + 3 x + 3 ) ^{2}}

求解步骤

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

使用指数法则:

\frac{1}{a} = a^{- 1}

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

使用连锁律:

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

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

\frac{d}{d x} ((x^{2} - 1) (x^{2} + 3 x + 3)) = 4 x^{3} + 9 x^{2} + 4 x - 3

= - \frac{1}{( ( x ^{2} - 1 ) ( x ^{2} + 3 x + 3 ) ) ^{2}} (4 x^{3} + 9 x^{2} + 4 x - 3)

化简

- \frac{1}{( ( x ^{2} - 1 ) ( x ^{2} + 3 x + 3 ) ) ^{2}} (4 x^{3} + 9 x^{2} + 4 x - 3) : - \frac{4 x ^{3} + 9 x ^{2} + 4 x - 3}{( x ^{2} - 1 ) ^{2} ( x ^{2} + 3 x + 3 ) ^{2}}

= - \frac{4 x ^{3} + 9 x ^{2} + 4 x - 3}{( x ^{2} - 1 ) ^{2} ( x ^{2} + 3 x + 3 ) ^{2}}

\int sin^{3} (7 x) cos^{- 2} (7 x) d x

t y^{^{'}} = 2 y + 3 t^{3} cos (2 t), y (\frac{π}{4}) = 0

d er i v a t i v e \frac{6 x ^{2} + 4 x + 6}{x}

\frac{\partial}{\partial y} (\frac{( x + y )}{z})

\frac{d}{d x} (\frac{2}{3} (- 2 e^{- 2 x} + e^{x}))