derivative of (8x^3-2x+9/x)

解答

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

\frac{16 x ^{3} - 9}{x ^{2}}

求解步骤

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

使用微分除法定则:

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

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

\frac{d}{d x} (8 x^{3} - 2 x + 9) = 24 x^{2} - 2

\frac{d x}{d x} = 1

= \frac{( 24 x ^{2} - 2 ) x - 1 \cdot ( 8 x ^{3} - 2 x + 9 )}{x ^{2}}

化简

\frac{( 24 x ^{2} - 2 ) x - 1 \cdot ( 8 x ^{3} - 2 x + 9 )}{x ^{2}} : \frac{16 x ^{3} - 9}{x ^{2}}

= \frac{16 x ^{3} - 9}{x ^{2}}

\frac{d}{d x} (\frac{1}{4 x ^{3} + 5 x ^{2} - 7 x + 8})

\int x (x^{2} + 3)^{7} d x

\frac{d}{d x} (x^{2} e n (x))

d er i v a t i v e y = tan (3 x)

\int \frac{1}{( x - 7 ) ^{2}} - 1 d x