derivative of (x^2+x+1/(x^2+3))

解

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

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

解答ステップ

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

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

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

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

拡張

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

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