((2t)/(t^2+1))^'

解

(\frac{2 t}{t ^{2} + 1})^{^{'}}

\frac{2 ( - t ^{2} + 1 )}{( t ^{2} + 1 ) ^{2}}

解答ステップ

(\frac{2 t}{t ^{2} + 1})^{'}

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 (\frac{t}{t ^{2} + 1})^{^{'}}

商の法則を適用:

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

= 2 \cdot \frac{t ^{^{'}} ( t ^{2} + 1 ) - ( t ^{2} + 1 ) ^{^{'}} t}{( t ^{2} + 1 ) ^{2}}

t^{'} = 1

(t^{2} + 1)^{'} = 2 t

= 2 \cdot \frac{1 \cdot ( t ^{2} + 1 ) - 2 tt}{( t ^{2} + 1 ) ^{2}}

簡素化

2 \cdot \frac{1 \cdot ( t ^{2} + 1 ) - 2 tt}{( t ^{2} + 1 ) ^{2}} : \frac{2 ( - t ^{2} + 1 )}{( t ^{2} + 1 ) ^{2}}

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