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

解

導関数

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

\frac{t ^{2} + 6 t - 12}{( t + 3 ) ^{2}}

解答ステップ

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

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

\frac{d}{d t} (t + 3) = 1

= \frac{( 2 t - 4 ) ( t + 3 ) - 1 \cdot ( t ^{2} - 4 t )}{( t + 3 ) ^{2}}

簡素化

\frac{( 2 t - 4 ) ( t + 3 ) - 1 \cdot ( t ^{2} - 4 t )}{( t + 3 ) ^{2}} : \frac{t ^{2} + 6 t - 12}{( t + 3 ) ^{2}}

= \frac{t ^{2} + 6 t - 12}{( t + 3 ) ^{2}}