derivative f(x)=(x^3+2x+7)/(x^2-5)

解

導関数

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

\frac{x ^{4} - 17 x ^{2} - 14 x - 10}{( x ^{2} - 5 ) ^{2}}

解答ステップ

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

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

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

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

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

拡張

(3 x^{2} + 2) (x^{2} - 5) - 2 x (x^{3} + 2 x + 7) : x^{4} - 17 x^{2} - 14 x - 10

= \frac{x ^{4} - 17 x ^{2} - 14 x - 10}{( x ^{2} - 5 ) ^{2}}