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

解答

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

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

求解步骤

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

化简

\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 )}

\frac{d}{d x} (39 arctan (x))

x \to - 5 + lim (\frac{x + 6}{x + 5})

d er i v a t i v e f (x) = \frac{1}{x} - 1

\int 15 tan^{3} (θ) d θ

x^{2} \frac{d y}{d x} + x y = 5