derivative of (8x/(1-tan(x)))

解答

\frac{d}{d x} (\frac{8 x}{1 - tan ( x )})

\frac{8 ( 1 - tan ( x ) + x sec ^{2} ( x ) )}{( 1 - tan ( x ) ) ^{2}}

求解步骤

\frac{d}{d x} (\frac{8 x}{1 - tan ( x )})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 8 \frac{d}{d x} (\frac{x}{1 - tan ( x )})

使用微分除法定则:

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

= 8 \cdot \frac{\frac{d x}{d x} ( 1 - tan ( x ) ) - \frac{d}{d x} ( 1 - tan ( x ) ) x}{( 1 - tan ( x ) ) ^{2}}

\frac{d x}{d x} = 1

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

= 8 \cdot \frac{1 \cdot ( 1 - tan ( x ) ) - ( - sec ^{2} ( x ) ) x}{( 1 - tan ( x ) ) ^{2}}

化简

8 \cdot \frac{1 \cdot ( 1 - tan ( x ) ) - ( - sec ^{2} ( x ) ) x}{( 1 - tan ( x ) ) ^{2}} : \frac{8 ( 1 - tan ( x ) + x sec ^{2} ( x ) )}{( 1 - tan ( x ) ) ^{2}}

= \frac{8 ( 1 - tan ( x ) + x sec ^{2} ( x ) )}{( 1 - tan ( x ) ) ^{2}}

\int_{0}^{4} 0.5 e^{- 0.5 t} d t

\int 8 sin^{3} (x) cos^{7} (x) d x

\int x^{5} e^{1 - x^{6}} d x

\int_{0}^{1} \frac{( t ^{2} )}{1 + t ^{6}} d t

\int (1 - cos (θ))^{2} d θ