((x^2+x+4)/(x+1))^'

解

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

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

解答ステップ

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

商の法則を適用:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

(x^{2} + x + 4)^{'} = 2 x + 1

(x + 1)^{'} = 1

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

簡素化

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

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