derivative y=(x^2-2x+5)/(x^2+2x-3)

解

導関数

y = \frac{x ^{2} - 2 x + 5}{x ^{2} + 2 x - 3}

\frac{4 x ^{2} - 16 x - 4}{( x ^{2} + 2 x - 3 ) ^{2}}

解答ステップ

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

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

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

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

拡張

(2 x - 2) (x^{2} + 2 x - 3) - (2 x + 2) (x^{2} - 2 x + 5) : 4 x^{2} - 16 x - 4

= \frac{4 x ^{2} - 16 x - 4}{( x ^{2} + 2 x - 3 ) ^{2}}