derivative of ((1-sec(x))/(tan(x)))

解答

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

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

求解步骤

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

使用微分除法定则:

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

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

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

化简

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

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

s im pl i f y x + \frac{1}{x ^{2}} + x^{3}

\int 5 e^{- 2 x} + 4 d x

6 \frac{d y}{d x} - y^{2} = 0

l a pl a ce t r an s f or m 2 cos (t)

2 x^{2} y^{^{''}} + 5 x y^{^{'}} + y = x^{2} - x