derivative (t^3+3t)/(t^2-4t+3)

解

導関数

\frac{t ^{3} + 3 t}{t ^{2} - 4 t + 3}

\frac{t ^{4} - 8 t ^{3} + 6 t ^{2} + 9}{( t ^{2} - 4 t + 3 ) ^{2}}

解答ステップ

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

商の法則を適用:

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

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

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

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

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

拡張

(3 t^{2} + 3) (t^{2} - 4 t + 3) - (2 t - 4) (t^{3} + 3 t) : t^{4} - 8 t^{3} + 6 t^{2} + 9

= \frac{t ^{4} - 8 t ^{3} + 6 t ^{2} + 9}{( t ^{2} - 4 t + 3 ) ^{2}}