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人気のある微分積分 >

derivative 5*2^{sqrt(t-2)}

解

導関数

5 \cdot 2^{t - 2}

解

\frac{5 ln ( 2 ) \cdot 2 ^{t - 2 - 1}}{t - 2}

解答ステップ

\frac{d}{d t} (5 \cdot 2^{t - 2})

定数を除去:

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

= 5 \frac{d}{d t} (2^{t - 2})

指数の規則を適用する:

a^{b} = e^{b l n (a)}

2^{t - 2} = e^{t - 2 l n (2)}

= 5 \frac{d}{d t} (e^{t - 2 l n (2)})

連鎖法則を適用:

e^{t - 2 l n (2)} \frac{d}{d t} (t - 2 ln (2))

= e^{t - 2 l n (2)} \frac{d}{d t} (t - 2 ln (2))

\frac{d}{d t} (t - 2 ln (2)) = \frac{ln ( 2 )}{2 t - 2}

= 5 e^{t - 2 l n (2)} \frac{ln ( 2 )}{2 t - 2}

簡素化

5 e^{t - 2 l n (2)} \frac{ln ( 2 )}{2 t - 2} : \frac{5 ln ( 2 ) \cdot 2 ^{t - 2 - 1}}{t - 2}

= \frac{5 ln ( 2 ) \cdot 2 ^{t - 2 - 1}}{t - 2}

グラフ

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