derivative of ln^3(sec(2^{\sqrt[3]{x}}))

Lösung

\frac{d}{d x} (ln^{3} (sec (2^{3 x})))

\frac{ln ( 2 ) \cdot 2 ^{3 x} ln ^{2} ( sec ( 2 ^{3 x} ) ) tan ( 2 ^{3 x} )}{x ^{\frac{2}{3}}}

Schritte zur Lösung

\frac{d}{d x} (ln^{3} (sec (2^{3 x})))

Wende die Kettenregel an:

3 (ln (sec (2^{3 x})))^{2} \frac{d}{d x} (ln (sec (2^{3 x})))

= 3 (ln (sec (2^{3 x})))^{2} \frac{d}{d x} (ln (sec (2^{3 x})))

\frac{d}{d x} (ln (sec (2^{3 x}))) = \frac{ln ( 2 ) \cdot 2 ^{3 x} tan ( 2 ^{3 x} )}{3 x ^{\frac{2}{3}}}

= 3 (ln (sec (2^{3 x})))^{2} \frac{ln ( 2 ) \cdot 2 ^{3 x} tan ( 2 ^{3 x} )}{3 x ^{\frac{2}{3}}}

Vereinfache

3 ln^{2} (sec (2^{3 x})) \frac{ln ( 2 ) \cdot 2 ^{3 x} tan ( 2 ^{3 x} )}{3 x ^{\frac{2}{3}}} : \frac{ln ( 2 ) \cdot 2 ^{3 x} ln ^{2} ( sec ( 2 ^{3 x} ) ) tan ( 2 ^{3 x} )}{x ^{\frac{2}{3}}}

= \frac{ln ( 2 ) \cdot 2 ^{3 x} ln ^{2} ( sec ( 2 ^{3 x} ) ) tan ( 2 ^{3 x} )}{x ^{\frac{2}{3}}}

d er i v a t i v e 13 e^{x^{2}}

y^{^{'}} + 3 y = 4

- y^{^{'}} x + y = - 7 x

\frac{d}{d x} ((x - 2) (x - 3)^{2})

\int_{0}^{\frac{π}{6}} 3 sec^{2} (x) d x