What is the derivative of (8x/(1-tan(x))) ?

The derivative of (8x/(1-tan(x))) is (8(1-tan(x)+xsec^2(x)))/((1-tan(x))^2)

What is the first derivative of (8x/(1-tan(x))) ?

The first derivative of (8x/(1-tan(x))) is (8(1-tan(x)+xsec^2(x)))/((1-tan(x))^2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

(\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}}

Simplify

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}}

\frac{d y}{d x} = \frac{( - 3 x ^{2} + 2 y ^{2} )}{4 x y}

\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

What is the derivative of (8x/(1-tan(x))) ?
The derivative of (8x/(1-tan(x))) is (8(1-tan(x)+xsec^2(x)))/((1-tan(x))^2)
What is the first derivative of (8x/(1-tan(x))) ?
The first derivative of (8x/(1-tan(x))) is (8(1-tan(x)+xsec^2(x)))/((1-tan(x))^2)