derivative f(t)=t^{1/2}(13t+2)

解

導関数

f (t) = t^{\frac{1}{2}} (13 t + 2)

\frac{39 t + 2}{2 t ^{\frac{1}{2}}}

解答ステップ

\frac{d}{d t} (t^{\frac{1}{2}} (13 t + 2))

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t^{\frac{1}{2}}, g = 13 t + 2

= \frac{d}{d t} (t^{\frac{1}{2}}) (13 t + 2) + \frac{d}{d t} (13 t + 2) t^{\frac{1}{2}}

\frac{d}{d t} (t^{\frac{1}{2}}) = \frac{1}{2 t ^{\frac{1}{2}}}

\frac{d}{d t} (13 t + 2) = 13

= \frac{1}{2 t ^{\frac{1}{2}}} (13 t + 2) + 13 t^{\frac{1}{2}}

簡素化

\frac{1}{2 t ^{\frac{1}{2}}} (13 t + 2) + 13 t^{\frac{1}{2}} : \frac{39 t + 2}{2 t ^{\frac{1}{2}}}

= \frac{39 t + 2}{2 t ^{\frac{1}{2}}}