d/(dt)((s)((k^2+1)sqrt(t)))

解

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

\frac{s ( k ^{2} + 1 )}{2 t}

解答ステップ

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

定数を除去:

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

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

累乗根の規則を適用する:

a = a^{\frac{1}{2}}

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

乗の法則を適用:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

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

簡素化

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

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