derivative y=(7x^3-3x^2-9)/(x^8)

解答

导数

y = \frac{7 x ^{3} - 3 x ^{2} - 9}{x ^{8}}

\frac{- 35 x ^{3} + 18 x ^{2} + 72}{x ^{9}}

求解步骤

\frac{d}{d x} (\frac{7 x ^{3} - 3 x ^{2} - 9}{x ^{8}})

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} ( 7 x ^{3} - 3 x ^{2} - 9 ) x ^{8} - \frac{d}{d x} ( x ^{8} ) ( 7 x ^{3} - 3 x ^{2} - 9 )}{( x ^{8} ) ^{2}}

\frac{d}{d x} (7 x^{3} - 3 x^{2} - 9) = 21 x^{2} - 6 x

\frac{d}{d x} (x^{8}) = 8 x^{7}

= \frac{( 21 x ^{2} - 6 x ) x ^{8} - 8 x ^{7} ( 7 x ^{3} - 3 x ^{2} - 9 )}{( x ^{8} ) ^{2}}

化简

\frac{( 21 x ^{2} - 6 x ) x ^{8} - 8 x ^{7} ( 7 x ^{3} - 3 x ^{2} - 9 )}{( x ^{8} ) ^{2}} : \frac{- 35 x ^{3} + 18 x ^{2} + 72}{x ^{9}}

= \frac{- 35 x ^{3} + 18 x ^{2} + 72}{x ^{9}}

\frac{d}{d x} (cos^{2} (x) \cdot sin (x))

\int_{0}^{4} 6 x d x

t an g e n t f (x) = x^{\frac{1}{2}}, (25, 5)

\frac{d}{d x} (lo g_{2} (3 2 x + 1))

\frac{d}{d x} (f (4 x^{5})) = 8 x^{4}