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

解答

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

\frac{3 x ^{2} + 2 x - 27}{( x ^{2} + 9 ) ^{2}}

求解步骤

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

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

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

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

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

乘开

(2 x - 3) (x^{2} + 9) - 2 x (x^{2} - 3 x + 8) : 3 x^{2} + 2 x - 27

= \frac{3 x ^{2} + 2 x - 27}{( x ^{2} + 9 ) ^{2}}

\int 12 + 4 x^{2} d x

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

\int_{- 3}^{2} (- x^{2} - x + 6) d x

\int (3 y^{2} + 6 y) (y^{3} + 3 y^{2} + 1)^{\frac{2}{3}} d y

t an g e n t f (x) = x^{2} - 8