derivative of (x^3+3x/(x^2-4x+3))

解

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

\frac{x ^{4} - 8 x ^{3} + 6 x ^{2} + 9}{( x ^{2} - 4 x + 3 ) ^{2}}

解答ステップ

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

商の法則を適用:

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

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

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

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

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

拡張

(3 x^{2} + 3) (x^{2} - 4 x + 3) - (2 x - 4) (x^{3} + 3 x) : x^{4} - 8 x^{3} + 6 x^{2} + 9

= \frac{x ^{4} - 8 x ^{3} + 6 x ^{2} + 9}{( x ^{2} - 4 x + 3 ) ^{2}}