derivative f(x)=((tan(x)-1))/(sec(x))

Lösung

ableitung von

f (x) = \frac{( tan ( x ) - 1 )}{sec ( x )}

\frac{sec ^{2} ( x ) - tan ( x ) ( tan ( x ) - 1 )}{sec ( x )}

Schritte zur Lösung

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

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( tan ( x ) - 1 ) sec ( x ) - \frac{d}{d x} ( sec ( x ) ) ( tan ( x ) - 1 )}{( sec ( x ) ) ^{2}}

\frac{d}{d x} (tan (x) - 1) = sec^{2} (x)

\frac{d}{d x} (sec (x)) = sec (x) tan (x)

= \frac{sec ^{2} ( x ) sec ( x ) - sec ( x ) tan ( x ) ( tan ( x ) - 1 )}{( sec ( x ) ) ^{2}}

Vereinfache

\frac{sec ^{2} ( x ) sec ( x ) - sec ( x ) tan ( x ) ( tan ( x ) - 1 )}{sec ^{2} ( x )} : \frac{sec ^{2} ( x ) - tan ( x ) ( tan ( x ) - 1 )}{sec ( x )}

= \frac{sec ^{2} ( x ) - tan ( x ) ( tan ( x ) - 1 )}{sec ( x )}

\int_{0}^{2} x^{2} d x

\frac{d}{d x} (\frac{2}{e ^{x} + e ^{- x}})

\int \frac{( x ^{2} + 2 x + 3 )}{( x ^{3} + 3 x ^{2} + 9 x )} d x

\int_{0}^{2 π} sin^{2} (4 x) d x

\int \frac{e ^{x}}{3} d x