derivative (9-sec(x))/(tan(x))

Lösung

ableitung von

\frac{9 - sec ( x )}{tan ( x )}

\frac{- tan ^{2} ( x ) sec ( x ) - sec ^{2} ( x ) ( 9 - sec ( x ) )}{tan ^{2} ( x )}

Schritte zur Lösung

\frac{d}{d x} (\frac{9 - sec ( x )}{tan ( x )})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 9 - sec ( x ) ) tan ( x ) - \frac{d}{d x} ( tan ( x ) ) ( 9 - sec ( x ) )}{( tan ( x ) ) ^{2}}

\frac{d}{d x} (9 - sec (x)) = - sec (x) tan (x)

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

= \frac{( - sec ( x ) tan ( x ) ) tan ( x ) - sec ^{2} ( x ) ( 9 - sec ( x ) )}{( tan ( x ) ) ^{2}}

Vereinfache

\frac{( - sec ( x ) tan ( x ) ) tan ( x ) - sec ^{2} ( x ) ( 9 - sec ( x ) )}{tan ^{2} ( x )} : \frac{- tan ^{2} ( x ) sec ( x ) - sec ^{2} ( x ) ( 9 - sec ( x ) )}{tan ^{2} ( x )}

= \frac{- tan ^{2} ( x ) sec ( x ) - sec ^{2} ( x ) ( 9 - sec ( x ) )}{tan ^{2} ( x )}

\frac{\partial}{\partial t} (\frac{3 r + s}{t})

y^{^{''}} - 2 y^{^{'}} + 6 y = 0

\int 4 sin^{2} (x) cos^{3} (x) d x

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

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