What is the derivative of (2x/(1-tan(x))) ?

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

What is the first derivative of (2x/(1-tan(x))) ?

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

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

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

Solution steps

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

Take the constant out:

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

= 2 \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}}

= 2 \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)

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

Simplify

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

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

y = 0, y = - x + 5, y = x + 1

- \frac{s - 2}{2 ( s ^{2} - 2 s + 2 )}

\int \frac{2 x}{x ^{2} + 1} d x

(12 - x) y^{2} = x^{3}, (6, 6)

\int (4 x^{3} + 2 x y^{2}) d x

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