derivative h(x)=(x^2+4x+1)/(3x+5)

解

導関数

h (x) = \frac{x ^{2} + 4 x + 1}{3 x + 5}

\frac{3 x ^{2} + 10 x + 17}{( 3 x + 5 ) ^{2}}

解答ステップ

\frac{d}{d x} (\frac{x ^{2} + 4 x + 1}{3 x + 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 ^{2} + 4 x + 1 ) ( 3 x + 5 ) - \frac{d}{d x} ( 3 x + 5 ) ( x ^{2} + 4 x + 1 )}{( 3 x + 5 ) ^{2}}

\frac{d}{d x} (x^{2} + 4 x + 1) = 2 x + 4

\frac{d}{d x} (3 x + 5) = 3

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

拡張

(2 x + 4) (3 x + 5) - 3 (x^{2} + 4 x + 1) : 3 x^{2} + 10 x + 17

= \frac{3 x ^{2} + 10 x + 17}{( 3 x + 5 ) ^{2}}