derivative of (3x-4/(2x^2+5))

解

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

\frac{- 6 x ^{2} + 16 x + 15}{( 2 x ^{2} + 5 ) ^{2}}

解答ステップ

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

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

\frac{d}{d x} (2 x^{2} + 5) = 4 x

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

拡張

3 (2 x^{2} + 5) - 4 x (3 x - 4) : - 6 x^{2} + 16 x + 15

= \frac{- 6 x ^{2} + 16 x + 15}{( 2 x ^{2} + 5 ) ^{2}}