derivative of (x^{-2}+x^3(4x)^{-2})

解答

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

\frac{3 ( 1 + x ^{3} ) ^{2} ( - 2 + x ^{3} )}{16 x ^{9}} - \frac{( 1 + x ^{3} ) ^{3}}{8 x ^{9}}

求解步骤

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

使用微分乘法定则:

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

f = (x^{- 2} + x)^{3}, g = (4 x)^{- 2}

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

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

\frac{d}{d x} ((4 x)^{- 2}) = - \frac{1}{8 x ^{3}}

= 3 (\frac{1}{x ^{2}} + x)^{2} (- \frac{2}{x ^{3}} + 1) (4 x)^{- 2} + (- \frac{1}{8 x ^{3}}) (x^{- 2} + x)^{3}

化简

3 (\frac{1}{x ^{2}} + x)^{2} (- \frac{2}{x ^{3}} + 1) (4 x)^{- 2} + (- \frac{1}{8 x ^{3}}) (x^{- 2} + x)^{3} : \frac{3 ( 1 + x ^{3} ) ^{2} ( - 2 + x ^{3} )}{16 x ^{9}} - \frac{( 1 + x ^{3} ) ^{3}}{8 x ^{9}}

= \frac{3 ( 1 + x ^{3} ) ^{2} ( - 2 + x ^{3} )}{16 x ^{9}} - \frac{( 1 + x ^{3} ) ^{3}}{8 x ^{9}}

\int 3 tan^{5} (x) d x

\int (1 - 5 y) d y

in v erse l a pl a ce \frac{1}{s ^{2} - 3 s + 2}

d er i v a t i v e arctan (5 x^{2} - 1)

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} = 1 + 9 x^{2}