derivative e^{(1/t ln((5^t+13^t)/2))}

Lösung

ableitung von

e^{(\frac{1}{t} l n (\frac{5 ^{t} + 1 3 ^{t}}{2}))}

(\frac{5 ^{t} + 1 3 ^{t}}{2})^{\frac{1}{t}} - \frac{ln ( \frac{5 ^{t} + 1 3 ^{t}}{2} )}{t ^{2}} + \frac{ln ( 5 ) \cdot 5 ^{t} + ln ( 13 ) \cdot 1 3 ^{t}}{t ( 5 ^{t} + 1 3 ^{t} )}

Schritte zur Lösung

\frac{d}{d t} (e^{\frac{1}{t} l n (\frac{5 ^{t} + 1 3 ^{t}}{2})})

Vereinfache

e^{\frac{1}{t} l n (\frac{5 ^{t} + 1 3 ^{t}}{2})} : (\frac{5 ^{t} + 1 3 ^{t}}{2})^{\frac{1}{t}}

= \frac{d}{d t} ((\frac{5 ^{t} + 1 3 ^{t}}{2})^{\frac{1}{t}})

Wende Exponentenregel an:

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

(\frac{5 ^{t} + 1 3 ^{t}}{2})^{\frac{1}{t}} = e^{\frac{1}{t} l n (\frac{5 ^{t} + 1 3 ^{t}}{2})}

= \frac{d}{d t} (e^{\frac{1}{t} l n (\frac{5 ^{t} + 1 3 ^{t}}{2})})

Wende die Kettenregel an:

e^{\frac{1}{t} l n (\frac{5 ^{t} + 1 3 ^{t}}{2})} \frac{d}{d t} (\frac{1}{t} ln (\frac{5 ^{t} + 1 3 ^{t}}{2}))

= e^{\frac{1}{t} l n (\frac{5 ^{t} + 1 3 ^{t}}{2})} \frac{d}{d t} (\frac{1}{t} ln (\frac{5 ^{t} + 1 3 ^{t}}{2}))

\frac{d}{d t} (\frac{1}{t} ln (\frac{5 ^{t} + 1 3 ^{t}}{2})) = - \frac{ln ( \frac{5 ^{t} + 1 3 ^{t}}{2} )}{t ^{2}} + \frac{ln ( 5 ) \cdot 5 ^{t} + ln ( 13 ) \cdot 1 3 ^{t}}{t ( 5 ^{t} + 1 3 ^{t} )}

= e^{\frac{1}{t} l n (\frac{5 ^{t} + 1 3 ^{t}}{2})} - \frac{ln ( \frac{5 ^{t} + 1 3 ^{t}}{2} )}{t ^{2}} + \frac{ln ( 5 ) \cdot 5 ^{t} + ln ( 13 ) \cdot 1 3 ^{t}}{t ( 5 ^{t} + 1 3 ^{t} )}

e^{\frac{1}{t} l n (\frac{5 ^{t} + 1 3 ^{t}}{2})} = (\frac{5 ^{t} + 1 3 ^{t}}{2})^{\frac{1}{t}}

= (\frac{5 ^{t} + 1 3 ^{t}}{2})^{\frac{1}{t}} - \frac{ln ( \frac{5 ^{t} + 1 3 ^{t}}{2} )}{t ^{2}} + \frac{ln ( 5 ) \cdot 5 ^{t} + ln ( 13 ) \cdot 1 3 ^{t}}{t ( 5 ^{t} + 1 3 ^{t} )}

\frac{d y}{d t} + 50 y = 0

a re a 9 x^{2} ln (x), 36 ln (x)

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\frac{\partial}{\partial v} (2 u)