derivative arcsec(1/(4t^5))

Lösung

ableitung von

arcsec (\frac{1}{4 t ^{5}})

- \frac{20 t ^{4}}{1 - 16 t ^{10}}

Schritte zur Lösung

\frac{d}{d t} (arcsec (\frac{1}{4 t ^{5}}))

Wende die Kettenregel an:

\frac{1}{( \frac{1}{4 t ^{5}} ) ^{2} ( \frac{1}{4 t ^{5}} ) ^{2} - 1} \frac{d}{d t} (\frac{1}{4 t ^{5}})

= \frac{1}{( \frac{1}{4 t ^{5}} ) ^{2} ( \frac{1}{4 t ^{5}} ) ^{2} - 1} \frac{d}{d t} (\frac{1}{4 t ^{5}})

\frac{d}{d t} (\frac{1}{4 t ^{5}}) = - \frac{5}{4 t ^{6}}

= \frac{1}{( \frac{1}{4 t ^{5}} ) ^{2} ( \frac{1}{4 t ^{5}} ) ^{2} - 1} (- \frac{5}{4 t ^{6}})

Vereinfache

\frac{1}{( \frac{1}{4 t ^{5}} ) ^{2} ( \frac{1}{4 t ^{5}} ) ^{2} - 1} (- \frac{5}{4 t ^{6}}) : - \frac{20 t ^{4}}{1 - 16 t ^{10}}

= - \frac{20 t ^{4}}{1 - 16 t ^{10}}

n = 0 \sum \infty \frac{z ^{n}}{n ^{2}}

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