derivative of ((2x^3+7(3x^2-5))/(x^2+1))

解

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

\frac{18 x ^{6} + 20 x ^{4} - 30 x ^{2} + 112 x}{( x ^{2} + 1 ) ^{2}}

解答ステップ

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

商の法則を適用:

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

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

\frac{d}{d x} ((2 x^{3} + 7) (3 x^{2} - 5)) = 30 x^{4} - 30 x^{2} + 42 x

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

= \frac{( 30 x ^{4} - 30 x ^{2} + 42 x ) ( x ^{2} + 1 ) - 2 x ( 2 x ^{3} + 7 ) ( 3 x ^{2} - 5 )}{( x ^{2} + 1 ) ^{2}}

拡張

(30 x^{4} - 30 x^{2} + 42 x) (x^{2} + 1) - 2 x (2 x^{3} + 7) (3 x^{2} - 5) : 18 x^{6} + 20 x^{4} - 30 x^{2} + 112 x

= \frac{18 x ^{6} + 20 x ^{4} - 30 x ^{2} + 112 x}{( x ^{2} + 1 ) ^{2}}