derivative of (sec(x+tan(x))^{-3})

Lösung

\frac{d}{d x} ((sec (x) + tan (x))^{- 3})

\frac{3}{( sec ( x ) + tan ( x ) ) ^{4} ( - 1 + sin ( x ) )}

Schritte zur Lösung

\frac{d}{d x} ((sec (x) + tan (x))^{- 3})

Wende die Kettenregel an:

- \frac{3}{( sec ( x ) + tan ( x ) ) ^{4}} \frac{d}{d x} (sec (x) + tan (x))

= - \frac{3}{( sec ( x ) + tan ( x ) ) ^{4}} \frac{d}{d x} (sec (x) + tan (x))

\frac{d}{d x} (sec (x) + tan (x)) = - \frac{1}{- 1 + sin ( x )}

= - \frac{3}{( sec ( x ) + tan ( x ) ) ^{4}} (- \frac{1}{- 1 + sin ( x )})

Vereinfache

- \frac{3}{( sec ( x ) + tan ( x ) ) ^{4}} (- \frac{1}{- 1 + sin ( x )}) : \frac{3}{( sec ( x ) + tan ( x ) ) ^{4} ( - 1 + sin ( x ) )}

= \frac{3}{( sec ( x ) + tan ( x ) ) ^{4} ( - 1 + sin ( x ) )}

\int_{0}^{4} 5 x d x

x \to 0 + lim (ln (tan (x)))

\frac{d}{d x} (arcsec (x) - 7 x)

\int \frac{1}{2 y - y ^{2}} d y

t an g e n t y = 2 x^{3} - 5 x + 1, (- 1, 4)