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

Lösung

ableitung von

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

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

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

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

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

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

Vereinfache

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

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

\frac{d}{d x} (e^{\frac{1}{x}} x^{2} - 1)

\frac{\partial}{\partial x} (sin (x) y)

\int (sec (x))^{4} d x

\int 16 + 9 x^{2} d x

\frac{e ^{x} d x}{- ( e ^{x} + 1 ) ^{3}} = \frac{e ^{y} d y}{( e ^{y} + 1 ) ^{2}}