What is the ((2sin(x))/(1-2cos(x)))^' ?

The ((2sin(x))/(1-2cos(x)))^' is (2(cos(x)(1-2cos(x))-2sin^2(x)))/((1-2cos(x))^2)

What is the first ((2sin(x))/(1-2cos(x)))^' ?

The first ((2sin(x))/(1-2cos(x)))^' is (2(cos(x)(1-2cos(x))-2sin^2(x)))/((1-2cos(x))^2)

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(\frac{2 sin ( x )}{1 - 2 cos ( x )})^{^{'}}

\frac{2 ( cos ( x ) ( 1 - 2 cos ( x ) ) - 2 sin ^{2} ( x ) )}{( 1 - 2 cos ( x ) ) ^{2}}

Solution steps

(\frac{2 sin ( x )}{1 - 2 cos ( x )})^{'}

Take the constant out:

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

= 2 (\frac{sin ( x )}{1 - 2 cos ( x )})^{^{'}}

Apply the Quotient Rule:

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

= 2 \cdot \frac{( sin ( x ) ) ^{^{'}} ( 1 - 2 cos ( x ) ) - ( 1 - 2 cos ( x ) ) ^{^{'}} sin ( x )}{( 1 - 2 cos ( x ) ) ^{2}}

(sin (x))^{'} = cos (x)

(1 - 2 cos (x))^{'} = 2 sin (x)

= 2 \cdot \frac{cos ( x ) ( 1 - 2 cos ( x ) ) - 2 sin ( x ) sin ( x )}{( 1 - 2 cos ( x ) ) ^{2}}

Simplify

2 \cdot \frac{cos ( x ) ( 1 - 2 cos ( x ) ) - 2 sin ( x ) sin ( x )}{( 1 - 2 cos ( x ) ) ^{2}} : \frac{2 ( cos ( x ) ( 1 - 2 cos ( x ) ) - 2 sin ^{2} ( x ) )}{( 1 - 2 cos ( x ) ) ^{2}}

= \frac{2 ( cos ( x ) ( 1 - 2 cos ( x ) ) - 2 sin ^{2} ( x ) )}{( 1 - 2 cos ( x ) ) ^{2}}

What is the ((2sin(x))/(1-2cos(x)))^' ?
The ((2sin(x))/(1-2cos(x)))^' is (2(cos(x)(1-2cos(x))-2sin^2(x)))/((1-2cos(x))^2)
What is the first ((2sin(x))/(1-2cos(x)))^' ?
The first ((2sin(x))/(1-2cos(x)))^' is (2(cos(x)(1-2cos(x))-2sin^2(x)))/((1-2cos(x))^2)