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

解答

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

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

求解步骤

\frac{d}{d x} (\frac{9 - 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} ( 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}}

化简

\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 )}

\int (x^{3} + 5) d x

\int e^{\frac{t}{5}} d t

x \to 2 - lim (\frac{x ^{2}}{x - 2})

n = 1 \sum \infty n^{3} (1 + n^{4})

d er i v a t i v e f (x) = x \cdot e^{2 x - 4}, x = 2